number theory
greatest common divisor
number theory solved

Prove that $x^4-y^4=z^2\ ,\ (x,y)=1$ has no Not obvious answer in $\mathbb{Z}.$

Consider k^2 - 2x^2 k + y^4.  It must have integer roots.  Now, the product of the roots is y^4  while the sum is 2x^2.  Put the roots as $x^2 + d, x^2 - d$.  Then the product is $x^4 - d^2$ so that $x^4 + y^4 = d^2$.  (*)

Lemma.  If there exists a solution to (*) (x,y,d); then there exists another solution (a,b,c) with c < d.

$(x^2, y^2, d)$ is a pythagorean triple.  If they have a common factor, dividing by this factor yields a smaller pythag triple $(a^2,b^2,c) c<d.$
Suppose they have no common factor.  By theory of pythag triples, we can write x^2 = m^2 - n^2, y^2=2mn.  So n^2 + x^2 = m^2,
and from theory we can write n = 2rs, x = r^2 - s^2, m = r^2 + s^2.  Now m(n/2) = (2mn)/4 = (y/2)^2 which is a perfect square because they are even.  So the product of coprime integers (m, n/2) are a square, therefore they are a square.  Also, rs is a perfect square.  So $(r^2,s^2,m)$ is a smaller pythag triple with $m < d$.

So there is no nontrivial solution to (*).  Experimentation with trivial solutions yield that none of these trivial solutions yield non-trivial solutions to the original problem.

ThanX $\text{Singular}$.But I think you solve $x^4+y^4=z^2$ what about $x^4-y^4=z^2$

We have, by mod 4, x is must be odd. WLOG, assumex,y are coprime.
Case 1) If y is is even, then x^2-y^2 and x^2+y^2 are coprime so that each is itself a perfect square, which is impossble by FLT.
Case 2) If y is odd, then gcd of x^2+y^2 and x^2-y^2 is 2, so that each number divided by 2 is itself a perfect squares, which reduces to case 1), so there are no solutions in posiituve integers. Hence one of x,y,z must be zero from where its easy...
I think i made mistake somewhere ;)

Bomb


statement 1 : when an ideal gas is heated in a rigid non conducting container then pressure becomes double if the temperature  is doubled.

statement 2:both the frequency of collisions and momentum transferred per collision becomes root(2) times.

I think both are correct and reason is the correct explanation of assertion.
[b]Explanation:-[/b]

[b]Statement 1[/b] is correct because as volume is constant,pressure is directly proportional to temperature.

[b]Statement 2:[/b]As the temperature doubles,the velocity which is proportional to square root of temperature becomes $ \sqrt {2}$ times and hence the momentum transferred also becomes $ \sqrt {2}$ times.And as velocity becomes $ \sqrt {2}$ times,the collision frequency which is proportional to velocity also becomes $ \sqrt {2}$ times.Now pressure is given by momentum transferred per collision times the frequency of collisions.As both frequency and momentum become $ \sqrt {2}$ times,their product,i.e.,pressure becomes double the original value.


trigonometry limit geometry calculus integration logarithms function

I would say the Binet's Formula.

$F_n = \frac{1}{\sqrt{5}}\left[ \left( \frac{1+\sqrt{5}}{2} \right)^n - \left( \frac{1-\sqrt{5}}{2} \right)^n \right]$

$e^{i \theta} = \cos {\theta} + i \sin {\theta}$

[quote="Rodolphe2005"]Hello,
my mother would paint mathematics formulas and she asked me what are most beautiful formulas, so what are your's ?

There is $e^{i\pi}=-1, \sum_{k=1}^{+\infty} \frac{1}{k^2}=\frac{\pi}{6}, ...$

Thanks[/quote]

Paint math formulas? Are you kidding? ;)

what means "kidding" ? If it's "joke", it's not a joke.

\[1+1=2\]
That's what all other formulas are based upon.

[quote="236factorial"]My favorite (because I've been using it 10^4384 times): $(a+b)^2=a^2+2ab+b^2$
Not beautiful.[/quote]
Are you sure you used it that many times? You must be very old!!! :D  :P  ;)  :rotfl:

[EDIT]Hi, I am sorry, I forgot to put a formula. Hmm, I found this formula in Discover magazine. It is a physics equation 10th dimension universe. Too bad it's old equation because those scientists already have 11th dimension...

$L=\langle\Psi^\dagger|i\partial_\tau-H|\Psi\rangle+g\langle\Psi^\dagger|\Psi^2\rangle$

whao $\displaystyle %Error. "Product" is a bad command.
_{k=1}^{\infty}\frac{ p_k^2+1}{p_k^2-1} = \frac 52$ is very nice, what's its name ? Demonstration is elementary ?

[quote="Danbert"]It's the prime product formula, proved by Ramanujan.  I assume that it's difficult to prove :)[/quote]

Product of the first p primes?

sorry to say, but physics is my first love.
$\imath \hbar \frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m} \frac{\partial^2\psi}{\partial x^2} + V\psi$

Where Can I find formulas that contain $e$ and $\pi$ ?

[quote="birkof"]I know that website and I have found only three: stirlings formula, euler's formula and puasson integral. Are there more? I thinks there are but where can I find them?[/quote]

[url]http://mathworld.wolfram.com/PiFormulas.html[/url]   :huh:  :)

Thanks for the interesting link but there are not formulas that contain e

$n^2+(n)+(n+1)=(n+1)^2$ Not that beautiful. Babbitt's theorem. Actually MY theorem. Pretty obvious.

$e^{\frac{1}{e}} \ge x^{\frac{1}{x}}$ for positive numbers $x$.

[quote="iostream.h"]sorry to say, but physics is my first love.[/quote]
Physics does have some cool equations.  How about all four of Maxwell's?
$\oint_S \mathbf{D} \cdot d\mathbf{A} = \int_V \rho dV \\ \oint_S \mathbf{B} \cdot d\mathbf{A} = 0 \\ \oint_C \mathbf{E} \cdot d\mathbf{l} = - \ { d \over dt } \int_S \mathbf{B} \cdot d\mathbf{A} \\ \oint_C \mathbf{H} \cdot d\mathbf{l} = \int_S \mathbf{J} \cdot d \mathbf{A} + {d \over dt} \int_S \mathbf{D} \cdot d \mathbf{A}$
And let's not forget about Laplace:
${\partial^2 \varphi\over \partial x^2 } + {\partial^2 \varphi\over \partial y^2 } + {\partial^2 \varphi\over \partial z^2 } = 0$.

Having seen but never used these equations, I must admit that part of their perceived elegance (at least for me) are the symbols themselves and the fact that they look good even though I don't know what they mean.

what about this formula: [img]http://mathworld.wolfram.com/images/eps-gif/ParallelianTheorem_1000.gif[/img]
[img]http://mathworld.wolfram.com/images/equations/Parallelian/equation1.gif[/img] where  [img]http://mathworld.wolfram.com/images/equations/Parallelian/inline1.gif[/img] is the area of the whole triangle. For more information have a look at http://mathworld.wolfram.com/Parallelian.html

How about something in combinatrics:
Catalan number:
$C(n)=\frac{ {2n \choose n} }{n+1}$

One with slight practical use(try it if you need to do trigno without calculator and can't bear with pi^5 in taylor's expansion- I mean, (355/113)^5 can be terrifying to do by hand): from Mathsworld
$\cos{\frac{\pi}{2}x} \approx 1 - \frac{x^2}{x+(1-x)\sqrt{\frac{2-x}{3}}}$

Sigh, now that everyone know and use cauchy, no one cares about Fibonacci anymore:
$(a^2+b^2)(c^2+d^2) = (ac-bd)^2+(ad+bc)^2$

\[ \left( \sum\limits_{i\in S}a_i\right) ^{k}=\sum\limits_{i_1,...,i_k\in S^{k}}a_{i_1}\cdot...\cdot a_{i_k}  \]
and the continuous version:
\[ \left( \int\limits_{I} f(t)dt \right) ^{k} = \int\limits_{I^{k}}f(t_1)\cdot...\cdot f(t_k) dt_1...dt_k.  \]

Nice formula birkof!!!

Here's another: ${\lambda}^4+{\lambda}^3+{\lambda}^2+{\lambda}+1=0$

the quartic formula is beautiful

no, actually its just plain ugly but whatever :)

I like my signature

Also, a cool thing to do:
Draw pascals triangle (pretty big) and then make a dark circle on the even numbers and erase the odd numbers and see what you get.  It's better on a computer.


One-sixth of one-third of one-fourth of what number is equal to five?

$ \frac{1}{6}\times \frac{1}{3}\times\frac{1}{4}x\equal{}5\implies\frac{1}{72}x\equal{}5\implies x\equal{}\boxed{360}$.


I have so many interests. I study olympiad math, calculus, physics, astronomy, and I play the piano. There are so many things to do and of course I can't do everything. As a result I have lots of books that I started studying and stopped in the middle, and I am not very good at any of this subjects. I find impossible to concentrate at one subject because if I study the same thing for much time I start to feel bored and want to study other things. So I jump from subject to subject and don't study anything deeply.

Can you give me some advice on how to find what is my favorite subject and pursue it deeply?  :)

Only you can know which is your favorite.  Many people have multiple interests, and they pursue them.  I don't know how many pursue 4-5 interests but many people go after math and a musical instrument.  Once you find which 2-3 you enjoy most, you'll find it easier to concentrate on them, and then you'll have fun and reap the benefits.

THe subjects are all closely related anyway, so I think jumping from subject to subject when you get bored won't be much of a problem.

The classic example of someone that pursued all of his interests successfully is Douglas Hofstadter with [url=http://en.wikipedia.org/wiki/G%C3%B6del,_Escher,_Bach]Godel, Escher, Bach[/url].

Music and math are actually very related. My english teacher was talking about this a few days ago and she said that a study showed that nearly all talented musical instrument players were profficient in math. (Well, she only mentioned it because we were reading a short reading - "A Middle-Class Wife").

[quote="Erken"][quote="JRav"] but many people go after math and a musical instrument[/quote]
Hmm...It happens quite often, when someone is a mathematician and a musician simultaneously,I am the happy exception,probably because I have never played on a musical instrument.
I wonder why a such things happen?I mean, what do these,it would seem,different activities, have in common?[/quote]

Math can be thought of as an art (as in Lockhart's Lament), a category to which music is classified; and music theory is in fact rather mathematical if you go deep enough - most do not realize the mathematics that goes behind tuning (even just the beginning level theory involving time signatures and circle of fifths can be mathematically stimulating).

And Sashsiam_2, of course this a very unbiased and impartial recommendation  :D , but I would focus the most on math and music. There are the extremes of the spectrum, and the purest of all studies, if you will. Physics and astronomy are too real to pursue. For example, if a theory in physics relies on a difficult mathematical result, the physicists will assume it's true and move on, and let the mathematicians work out the details and justification (they say mathematicians rather behind the physicists currently - isn't that so mean of the physicists?  :lol: ). And astronomy may at first strike at the imagination with wonder at the size of the universe, but math has an infinity of infinities still greater, and plus you don't need a lot of $ \$\$\$$ to study math. (Study math = book, class, study science = book, class, kabajillion dollar particle accelerator thingymabober)


What is the number of square units in the area of trapezoid ABCD with vertices A(0,0), B(0,-2), C(4,0), and D(4,6)?

$ \frac{(AB\plus{}CD) \times AC}{2} \equal{} \frac{(2\plus{}6) \times 4}{2} \equal{} \boxed{16}$.

or you could shoelace it.

Listing them in order:
\[ (0,0)\newline(0,\minus{}2)\newline(4,0)\newline(4,6)\newline(0,0)\]

Then we pair up the coordinates.
$ \frac{1}{2}|(0\times\minus{}2\plus{}0\times0\plus{}4\times6\plus{}4\times0)\minus{}(0\times0\plus{}\minus{}2\times4\plus{}0\times4\plus{}6\times4)|\equal{}\frac{1}{2}|24\plus{}8|\equal{}\frac{1}{2}\times32\equal{}\boxed{16}$.

If you don't understand some of the steps I did, look up the [url=http://www.artofproblemsolving.com/Wiki/index.php/Shoelace_Theorem]shoelace theorem[/url].


I have a small Log problem that has been bothering me lately...Maybe you guys can decipher this.

SOLVE for t.
The e is e as in Euler so...think natural log (ln).

A = B[1-e^(t/RC)] + C[e^(t/RC)]

BTW, I'm new here. HOLLA!

$ A \equal{} B(1\minus{}e^{t/RC}) \plus{} Ce^{t/RC} \equal{} B \plus{} (C\minus{}B) e^{t/RC}$

$ \frac{A\minus{}B}{C\minus{}B} \equal{} e^{t/RC}$

$ \frac{t}{RC} \equal{} \ln \left( \frac{A\minus{}B}{C\minus{}B} \right)$

$ t \equal{} RC \cdot \ln \left( \frac{A\minus{}B}{C\minus{}B} \right)$

[quote="TZF"]$ A \equal{} B(1 \minus{} e^{t/RC}) \plus{} Ce^{t/RC} \equal{} B \plus{} (C \minus{} B) e^{t/RC}$

$ \frac {A \minus{} B}{C \minus{} B} \equal{} e^{t/RC}$

$ \frac {t}{RC} \equal{} \ln \left( \frac {A \minus{} B}{C \minus{} B} \right)$

$ t \equal{} RC \cdot \ln \left( \frac {A \minus{} B}{C \minus{} B} \right)$[/quote]

Thanks...Really appreciate it...But I don't understand how you got from the first step to the second.

$ A \equal{} B(1 \minus{} e^{t/RC}) \plus{} Ce^{t/RC} \equal{} B \plus{} (C \minus{} B) e^{t/RC}$
$ A\minus{}B\equal{}(C \minus{} B) e^{t/RC}$
$ \frac {A \minus{} B}{C \minus{} B} \equal{} e^{t/RC}$

[quote="math154"]$ A \equal{} B(1 \minus{} e^{t/RC}) \plus{} Ce^{t/RC} \equal{} B \plus{} (C \minus{} B) e^{t/RC}$
$ A \minus{} B \equal{} (C \minus{} B) e^{t/RC}$
$ \frac {A \minus{} B}{C \minus{} B} \equal{} e^{t/RC}$[/quote]

I don't get how...$ B(1 \minus{} e^{t/RC}) \plus{} Ce^{t/RC}$
turns into
$ B \plus{} (C \minus{} B) e^{t/RC}$

Someone please break that down.

$ B(1 \minus{} e^{t/RC}) \plus{} Ce^{t/RC}$
$ B\minus{}Be^{t/RC}\plus{}Ce^{t/RC}$
$ B\plus{}(C\minus{}B)e^{t/RC}$

Ahhhhh....I finally get it now...Thanks duke...Appreciate it to death.

So unless I'm mistaken this equation comes from the derivation of the RC time constant in a resistor-capacitor circuit.

Am I rusty on physics, or is that right-ish?

I forgot the exact name...But it's actually the formula for Electrical Circuit...Or something like that.
The A, B and C...were actually Vs, V sub 0, and V sub (I forget)
I substituted it to make it easier.


Suppose $ x\not \equal{} y$ and $ x,y\in \mathbb {N}$

Let $ A ,G$ be the $ A.M.$ and $ G.M.$ for $ x,y$ respectively.

If $ A$ and $ G$  both are double-digit numbers ,
and their tenth and unit digits switch  , then $ A$ and $ G$ will be exchanged each other.

Find $ x, y.$

 :P

[quote="vinskman"]Suppose $ x\not \equal{} y$ and $ x,y\in \mathbb {N}$

Let $ A ,G$ be the $ A.M.$ and $ G.M.$ for $ x,y$ respectively.

If $ A$ and $ G$  both are double-digit numbers ,
and their tenth and unit digits switch  , then $ A$ and $ G$ will be exchanged each other.

Find $ x, y.$

 [/quote]

Anybody has idea?
 :wink:

Well I have somewhat of an idea..

By the AM-GM inequality we have:

$ \frac{x \plus{} y}{2} > \sqrt{xy}$

(We know that they are not equal since x and y are not equal)

Since the digits of the AM can be reversed to make the GM let's let the left side equal AB and the right side equal BA.

$ AB > BA$

This implies that A > B.

Maybe you can go on from here?

Since $ \sqrt {xy} \equal{} G$, $ xy \equal{} G^2$. Similarly, since $ \frac {x \plus{} y}{2} \equal{} A$, $ x \plus{} y \equal{} 2A$. Solving for $ x$ and $ y$, we get $ A\pm{\sqrt {A^2 \minus{} G^2}}$. Hence if $ (a,b,c)$ are pythagorean triples such that  $ a$, $ b$, and $ c$ are all two digit numbers, there are two solutions for $ x$ and $ y$, namely $ (c \minus{} a,c \plus{} a)$ and $ (c \minus{} b,c \plus{} b)$. So $ A \equal{} 65$, and $ G \equal{} 56$, and $ x \equal{} 32$, $ y \equal{} 98$.

[hide="wrong solution  :D "]
Let's say that $ \frac {x \plus{} y}{2} \equal{} 10a \plus{} b \cdots \alpha$ and $ \sqrt {xy} \equal{} a \plus{} 10b \cdots \beta$

Then, we have $ 10\alpha \minus{} \beta \equal{} 5(x \plus{} y) \minus{} \sqrt {xy} \equal{} 100a \plus{} 10b \minus{} a \minus{} 10b \equal{} 99a$

Take a look at $ \sqrt {xy}.$ Since $ x$ and $ y$ are both two-digit integers, maximum value of $ \sqrt {xy}$ is 99.

First, we have to find how many value can $ x \plus{} y$ have.

Obtimizing, we have $ 5(x \plus{} y) \equal{} 99(1 \plus{} a) \implies a \equal{} 4, 9$

If $ a \equal{} 4$, $ x \plus{} y \equal{} 99$, which can't be true because $ \frac {x \plus{} y}{2}$ is an integer.

If $ a \equal{} 9$, $ x \equal{} y \equal{} 99$. Then, we have $ \sqrt {xy}$=99, which also can't be true because $ a \equal{} b \equal{} 9$.

Thus, there is no answer.[/hide]

Twin Prime Conjecture: $ 32\times 98\neq 33^2$

[quote="FantasyLover"][hide="solution"]
Let's say that $ \frac {x \plus{} y}{2} \equal{} 10a \plus{} b \cdots \alpha$ and $ \sqrt {xy} \equal{} a \plus{} 10b \cdots \beta$

Then, we have $ 10\alpha \minus{} \beta \equal{} 5(x \plus{} y) \minus{} \sqrt {xy} \equal{} 100a \plus{} 10b \minus{} a \minus{} 10b \equal{} 99a$

Take a look at $ \sqrt {xy}.$ Since $ x$ and $ y$ are both two-digit integers, maximum value of $ \sqrt {xy}$ is 99.

First, we have to find how many value can $ x \plus{} y$ have.

Obtimizing, we have $ 5(x \plus{} y) \equal{} 99(1 \plus{} a) \implies a \equal{} 4, 9$

If $ a \equal{} 4$, $ x \plus{} y \equal{} 99$, which can't be true because $ \frac {x \plus{} y}{2}$ is an integer.

If $ a \equal{} 9$, $ x \equal{} y \equal{} 99$. Then, we have $ \sqrt {xy}$=99, which also can't be true because $ a \equal{} b \equal{} 9$.

Thus, there is no answer.[/hide]

Twin Prime Conjecture: $ 32\times 98\neq 33^2$[/quote]

my bad, $ 33$ is supposed to be $ 56$, I edited it, thanks for pointing out.

The following is a detailed explanation of what would otherwise be a fast guess-and check method.

[hide="Systematic Search"]AM-GM must be a multiple of 9 here. Set up the equation:

$ \frac {x \plus{} y}{2} \minus{} \sqrt {xy} \equal{} 9 N \implies \left( \sqrt {x} \minus{} \sqrt {y} \right) ^2 \equal{} 18 N$

The easiest case to treat is $ N \equal{} 1$. We have $ \sqrt {x} \minus{} \sqrt {y} \equal{} 3 \sqrt {2}$, so we decide that $ x \equal{} 2 (v \plus{} 3)^2$ and $ y \equal{} 2 v^2$. Their respective AM and GM are $ v^2 \plus{} (v \plus{} 3)^2$ and $ 2v(v \plus{} 3)$. We easily find that $ v \equal{} 4$ generates the only valid integer solution, so $ (x,y) \equal{} (98,32)$, and their AM/GM's are $ 65, \; 56$

Another really easy case to treat is $ N \equal{} 2$. We have $ \sqrt {x} \minus{} \sqrt {y} \equal{} 6$, an integer, so both $ x$ and $ y$ must be perfect squares. We check pairs of squares whose squareroots differ by 6 and find no solutions.

For $ N \equal{} 3$, we find $ \sqrt {x} \minus{} \sqrt {y} \equal{} 3 \sqrt {6}$. Thus, we conclude that $ x \equal{} 6 (v \plus{} 3)^2$ and $ y \equal{} 6 v^2$, whose AM and GM are $ 6v^2 \plus{} 18v \plus{} 27$ and $ 6v^2 \plus{} 18v$. $ AM \leq 100$, so $ v \leq \approx 2.2 \implies v \equal{} 1,2$. Neither generates an appropriate solution.

It's easy to see that we're pretty much hopeless now. If you wish to continue, it looks like this:

[hide]For $ N \equal{} 4$, we find $ \sqrt {x} \minus{} \sqrt {y} \equal{} 6 \sqrt {2}$. Thus, we conclude that $ x \equal{} 2 (v \plus{} 6)^2$ and $ y \equal{} 2 v^2$, whose AM and GM are $ 2v^2 \plus{} 12v \plus{} 36$ and $ 2v^2 \plus{} 12v$. $ AM \leq 100$, so $ v \leq \approx 3.4 \implies v \equal{} 1,2,3$. None generate an appropriate solution.

$ N \equal{} 5$ gives us the following expression:
$ \sqrt {x} \minus{} \sqrt {y} \equal{} 3 \sqrt {10} \; \implies \; x \equal{} 10 (v \plus{} 3)^2, \; \; y \equal{} 10 v^2$, whose AM is $ 10 v^2 \plus{} 30 \plus{} 45$. $ AM \leq 100 \implies v \leq \approx 1.3 \implies v \equal{} 1$, which doesn't work. We realize that for any more odd numbers, we'll just be in a worse situation (next up is $ 3 \sqrt {14}$ for $ N \equal{} 7$), so we give up with odd numbers. Let's see how the next even number fares:

For $ N \equal{} 6$, we find $ \sqrt {x} \minus{} \sqrt {y} \equal{} 6 \sqrt {3} \; \; \implies x \equal{} 3 (v \plus{} 6)^2, \; \; y \equal{} 3 v^2$, whose AM is $ 3 v^2 \plus{} 18v \plus{} 54$. $ AM \leq 100$, so $ v \leq \approx 1.9 \implies v \equal{} 1$, which doesn't work. We realize that we're in the same dunghole for the remaining even numbers, so we end our search.[/hide]
The unique solution is $ (x,y) \equal{} (98,32)$, with $ AM \equal{} 65$ and $ GM \equal{} 56$[/hide]


let $ a$,$ b$ and $ c$ positives reals numbers such that $ a\plus{}b\plus{}c\equal{}1$.
show that:
$ \frac{a^3}{a^2\plus{}b^2}\plus{}\frac{b^3}{b^2\plus{}c^2}\plus{}\frac{c^3}{c^2\plus{}b^2}\geq\frac{1}{2}$

\[ \frac{a^3}{a^2\plus{}b^2}\plus{}\frac{b^3}{b^2\plus{}c^2}\plus{}\frac{c^3}{c^2\plus{}b^2}\geq\frac{1}{2} \leftrightarrow \sum_{cyc}\frac{ab^2}{a^2\plus{}b^2} \le \frac{1}{2}\]
which is obviously true by am-gm; $ a^2\plus{}b^2 \ge 2ab$

$ \frac{a^3}{a^2\plus{}b^2}\geq a\minus{}\frac{b}{2}\Leftrightarrow b(a\minus{}b)^2\geq 0$
$ \Rightarrow\sum\frac{a^3}{a^2\plus{}b^2}\geq\frac{\sum a}{2}\equal{}\frac{1}{2}$


Given the points K = (0,0), L = (7,−1), M = (9,3), P = (6,7), Q = (10,5), and R = (1, 2), triangles KLM and RP Q are congruent. Find coordinates for the point in triangle PQR that corresponds to (6,2) in triangle KLM.

It seems that triange $ PQR$ can be got from triangle $ LKM$ by translating $ 1$ to the right and up $ 2$ and then reflecting in the line $ RQ$. This takes $ (6, 2)$ to $ (7, 4)$, which is on the line $ y\equal{}\frac{1}{3}x \plus{} \frac{5}{3}$, that is, the line $ RQ$.

[quote="AndrewTom"]It seems that triange $ PQR$ can be got from triangle $ LKM$ by translating $ 1$ to the right and up $ 2$ and then reflecting in the line $ RQ$. This takes $ (6, 2)$ to $ (7, 4)$, which is on the line $ y \equal{} \frac {1}{3}x \plus{} \frac {5}{3}$, that is, the line $ RQ$.[/quote]

I can't tell if that's right, but it might be.

How does one set up and lay out a problem like this -- in which one is given a pre-image and an image and is asked to "reverse engineer" the transformation -- in a way that's both clear and eliminates doubt as to its correctness?

Hi Peterhi.

I just sketched the two triangles on squared paper and then it was easy to see. You can check by taking other points (forwards or backwards).

Hope this helps; if not, I'll try again.

Thanks for the offer -- you've been more than helpful.


Do any of you know Forth? Can you recommend any good books/sites for learning Forth?

i learned forth some years ago but i think now this was an error: it's an old langage, it's quite long to write efficient program and it's not that readable ... i don,t remember why i learnt that ? and you ?

I know that it's an old language, and that it's not very popular (that's why there aren't that many books on it) but I just wanted to try it out.. Maybe it's the RPN that I'm attracted to, I don't know, but anyway, if you have any good tips/pointers/books/websites/etc then that'd be great.


What do you do to make other people laugh? What do they do to make you laugh?

Whenever I recite 'All your base', I never forget the 'Ha ha ha' right after the 'All your base are belong to us'.

i tell jokes that are so stupid that u have to laugh

Like: Yo mama so fat that she has to wear two watches cause she takes up two time zones?  :lol:

No offense. Anyone.


hAPPY NEW YEAR!!!!!!!!!!!!!

 :rotfl:

SO WHAT ARE YOUR NEW YEARS RESOLUTIONS?

One of mine is to always check if the Caps Lock is on...

[quote="Ignite168"]How about you make it to scan the first 10 topics or so before posting a double post?[/quote]

THE MORE HAPPY WISHES THE MORE MERRIER!!!

HAPPY NEW YEAR!!

I think restated, we would have:

"The more spam the merrier"

Seriously this is called games and FUN factory for a reason, not G&spam factory.

i was still officially first (look at my blog)

-jorian


Is there a decreasing bijection $ f: \mathbb{R}\rightarrow\mathbb{R}$ such that $ \forall x\in\mathbb{R}$
\[ f^{\minus{}1}(x^3)\leq f(f(x)^2\plus{}f(x))\]
If there is, find all such functions.

As $ x \to \minus{} \infty$, $ x^3 \to \minus{} \infty$ and so $ f^{ \minus{} 1}(x^3) \to \infty$.  Similarly, $ f(x) \to \infty$ and so $ f(x)^2 \plus{} f(x) \to \infty$ and thus $ f(f(x)^2 \plus{} f(x)) \to \minus{} \infty$.

(What an odd question.  Where did it come from?)


real analysis
superior algebra
superior algebra unsolved

Do there exist a subset $ A$ of $ \mathbb R^2$ such that $ 0 < \mu_L(A)$ , and every line intersect $ A$ at finitely
many points?

Allah does not accept any act without knowledge, there is no knowledge without act, so whoever knows, knowledge leads him to act, and whoever does not act gets no knowledge, but belief is a little of a little.[b][/b]

[quote="christ"]Do you mean lebesgue measure "u_L"  ?[/quote]

yes  :wink:

A brief solution :
There is at least one measurable set with the first condition :
A={(x,x^3) belong to R^2| x belong to R} but u_L(A) =0 which not satisfy the other condition .So we can talk about measurable sets such as A .
If m1 be Lebesgue measure space of R then m2 (Lebesgue measure space of R^2 )will be completion of m1 X m1 (Real and Complex Analasis ,Walter Rudin ,the third Edition, theorem 11.8).
Defining  product measure shows that u_1,1(A) =0 (measure associated to m1 X m1) .Then m2(A)=u_L(A)=0 ,so there is not A with these conditions  .


The diagonals of a trapezoid are perpendicular with lengths 7 and 9. Find the length of the median of the trapezoid.

Please explain what you are referring to as the median of the trapezoid... from what I understand, it is nonconstant.

The median of a trapezoid is usually defined as the segment connecting the midpoints of each leg

It is parallel to each base and is the average of their lengths


get x by itself.

a^2 x + (a-1) = (a +1)x

Your goal is to try to get x by itself.

$a^2x+a-1=x(a-1)$

$a^2x+a-1=ax+x$

$a^2x-ax-x=1-a$

$x(a^2-a-1)=1-a$

$x=\frac{1-a}{a^2-a-1}$

I don't get a thank you?  :(

Just kidding. But I would like a thank you for such a fast reply.  :(

he posted this in classroom and here. :?

[quote="4everwise"]He posted this in every forum. It was posted in Getting started, classroom, basics, and in the advanced caclculus section.  :mad:

Please don't do this again. :noo:[/quote] :ninja:  sry.. im a newb

Oh, sorry for not noticing that. I try to be a good moderator. I will now give a better explanation.

You want to get x by itself in some way. In this case, the best way to do that will be to take all the terms with x in them to one side and all the other terms to the other side.

First, multiply everything else so you can do that.
$a^2x+a-1=x(a-1)$
turns into
$a^2x+a-1=ax+x$

Now comes the relocation. All the stuff with x is on one side, everything else on the other.
$a^2x-ax-x=1-a$

This is probably the part where I lost you earlier. Because everything on the left side has an "x," you factor out the x.
$x(a^2-a-1)=1-a$

Because this is now multiplication, divide both sides by $(a^2-a-1)$ to get the answer
$x=\frac{1-a}{a^2-a-1}$

If you still don't understand, notify me and I'll try to explain in further detail.


Find all primes numbers $p,q,r$ such that $p+q=r+1$.

Sure this is not open¿ ;)
But there probably is a solution for all $r$ ;)


trigonometry
algebra
polynomial
induction
Irrational numbers

Show that $\tan \left(  \frac{\pi}{m} \right)$ is irrational for all positive integers $m \ge 5$.

We needs  the following
Lemma: If the polynomial
$ P(x) = a_0 x^n +a_1 x^{n-1}+ ...+ a_{n-1}x + a_n \in Z[x], a_0 \ne 0$
admits as a root an irreducible fraction $ \frac {a}{b} \ne 0$,
then $ a$ divides $ a_n$ and $ b$ divides $ a_0$.
Now let us consider two cases.

1° $ m = 2k+1, k\in Z$,

$ \tan(m \alpha )= \frac { \pm x ^{m} + P_m (x)}{1 + Q_m (x)}$ where
$ x =\tan \alpha$, and $ P_m (x)$, $ Q_m (x)$ are polynomials of integer coefficients, of order at most $ m -1$.
By induction,
for $ m=1$ it is obvious. Let us suppose that is true for some $ m$, and consider
$ \tan (m+2) \alpha =\tan ( m\alpha + 2 \alpha) = \frac{\tan m\alpha +\tan 2 \alpha}{1 - ( \tan m\alpha)( \tan 2 \alpha)} =
(\frac{\pm x + P_m (x)}{1 + Q_m(x)} +\frac{2x}{1 - x^2}) / W(x)=
\frac {\pm x^{m+2} \pm x^m + (1 - x^2) p_m (x) + 2x (1 + Q_m(x))}{1+ Q(x) - x^2(1+Q_m(x))-2x(\pm x\m + P_m(x))}$,
(where $ W(x) = 1 - 2x ( \frac{ \pm x^m +P_m(x)}{(1-x^2) (1+Q_m(x))} )$ ),
which  imply that the representation is valid for $ m+2$.
Now, let $ \alpha = \frac{ \pi}{m}$. It follows that $ \pm x_0^{m} + P_m (x_0) = 0$.

The lemme implies that $ x_0 = tan \alpha$ must be integer.
But we know, that $ 0 < \tan \alpha <1$, so contradiction.
2° $ m= 2^k \cdot n$, with $ n$ is odd. Since
$ \tan (2 \alpha) = 2 ( \frac {\tan \alpha}{1-  (\tan  \alpha)^2})$,
and supposing $ \ tan (\frac{ \pi}{ 2^k n} ) \in Q$ we have that all the terms of the following sequence
 $ \ tan (\frac{ \pi}{ 2^{k-1} n} )$, $ \tan (\frac { \pi}{ 2^{k-2} n} )$,.... $ \ tan (\frac{ \pi}{ 2^0 n} )$= $ \ tan (\frac{ \pi}{ n}$ )
are all rational. But that is contradiction.
So that is done


[quote="ice_age"]The area of a convex quadrilateral is $64\ m^{2}$, the sum of the lengths of a diagonal and a pair of opposite sides is $16\sqrt2$ m.
Find the length of the other diagonal.[/quote]

mlok, thanks!

To summarize, I have the following final solution:

Recall this inequality:     $a\cdot b\leq(\frac{a+b}{2})^{2}$.

Suppose the length of the known diagonal is $y$, then the sum of the two opposite sides is $(16\sqrt2-y)$.

When the two opposite sides are perpendicular to the diagonal, the area of the quadrilateral is the largest.
$S_{quadrilateral}\leq\frac{1}{2}y(16\sqrt2-y)\leq\frac{1}{2}\cdot(\frac{y+(16\sqrt2-y)}{2})^{2}=64$.
   iff $y=16\sqrt2-y$, i.e., $y=8\sqrt2$, the equality holds. And only under this condition, the area of the quadrilateral is 64.

Therefore, the length of the known diagonal is $8\sqrt2$, and the sum of the two opposite sides is also $8\sqrt2$.

Finally, we know that the other diagonal is $16 cm$ in length.

[quote="mlok"][hide="An idea"] Let $a$ and $c$ be two opposite sides and let $p$ be the diagonal such that \[a+c+p=16\sqrt{2}\] The diagonal $p$ splits the quadrilateral into two triangles. These trianges have base $p$ and heights at most $a$ and $c$. Therefore, their areas are at most $pa/2$ and $pc/2$. But the areas add up to $64$, hence \[p(a+c)\ge 128\] You still don't have enough information to find $a$ or $c$, but you can find $a+c$ and $p$, and then the other diagonal.

Hint: $x+1/x\ge 2$ for any $x>0$, with equality only for $x=1$.[/hide][/quote]

thanks.
and after further searching, I found that actually this question is similar to an IMO 1976 question.
see url below:
http://www.kalva.demon.co.uk/imo/imo76.html

Also copied here:
IMO 1976

Problem A1
A plane convex quadrilateral has area 32, and the sum of two opposite sides and a diagonal is 16. Determine all possible lengths for the other diagonal.

Solution
At first sight, the length of the other diagonal appears unlikely to be significantly constrained. However, a little experimentation shows that it is hard to get such a low value as 16. This suggests that 16 may be the smallest possible value.
If the diagonal which is part of the 16 has length x, then the area is the sum of the areas of two triangles base x, which is xy/2, where y is the sum of the altitudes of the two triangles. y must be at most (16 - x), with equality only if the two triangles are right-angled. But x(16 - x)/2 = (64 - (x - 8)2)/2 ≤ 32 with equality only iff x = 8. Thus the only way we can achieve the values given is with one diagonal length 8 and two sides perpendicular to this diagonal with lengths totalling 8. But in this case the other diagonal has length 8√2.

This is not just similar to, but [b]equivalent to[/b] [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=61042]IMO 1976 problem 1[/url] after the quadrilateral is stretched (homothetically) with factor $\sqrt2$.

  Darij


It appears that this is a popular question.
Posted [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=91594]here[/url] and [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=105367]here[/url] and [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=107345]here[/url] and [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=113745]here[/url].


The function $f(x)$ satisfies $f(2+x) = f(2-x)$ for all real numbers $x$. If the equation $f(x) = 0$ has exactly four distinct real roots, then the sum of these roots is

A. 0
B. 2
C. 4
D. 6
E. 8

$f(2+x)=f(-x+2)$
$f(2+(x+2)=f(-(x+2)+2)$
$f(x+4)=f(-x)$
$f(a)=0$
$f(b)=0$
$f(c)=0$
$f(d)=0$

If we substitute $-a, -b, -c, -d$ fo $x$ in our last equation, then we get:
$f(-a+4)=0$
$f(-b+4)=0$
$f(-c+4)=0$
$f(-d+4)=0$
Function has exactly four distinct roots, so set ${[a,b,c,d]}$is equal to set ${[-a+4, -b+4, -c+4, -d+4]}$

After considering a few cases we get without loosing generality:

$a=-b+4$
$b=-c+4$
$c=-d+4$
$d=-a+4$
So the sum of the roots is equal to $\boxed{8}$

it's really possible someone would choose 0, or more possibly 2.
why 2? because if it's f(x)=f(-x), aka f(0+x)=f(0-x), the result would have been 0. now we replace the 0 with 2, wouldn't the result be 2?! well, it's actually __(as mentioned above).
for a single problem, the mistake can be easily detected. But if this problem is only a small segment of a complicated big problem, and you get everything right except here, it's gonna be big trouble(which is what happened to me).

Lets take x1, a solution for f(x)=0. There is a real number y so that x1=y+2. so we have f(x1)=f(y+2)=f(2-y)=0, so 2-y is also a solution for f(x)=0, so x2=2-y, so x1+x2=4.
Now we can choose from one of the 2 roots left another root, x3. we follow the same steps as in the previous case and we prove x3+x4=4.
So x1+x2+x3+x4=8.

This function is symmetric by the point 2. And so, the roots are separeted by this center of symmetry. Suppose that $x_1,x_2$ are the "distances" from 2 that these roots are. And so, we have as roots $2-x_1,2-x_2$, since they're symmetric by 2, we know where the other roots are $2+x_1,2+x_2$. Summing them we find 8.

Isn't this fairly obvious? If $f(x) = 0$, let $x = 2 + y$ for some $y$; now let $x_0 = 2 - y$. We could do this for every solution to $f(x) = 0$, producing another solution, which, when summed with the original gives you 4.. So each 2 solutions represents a sum of 4 for, and being 4 solutions, the sum of all the roots must be 8.

\boxed{E}

if $r_1 = 2+a$ is a root, so is $r_2 = 2-a$ because they are symmetrical about $x=2$

$(2-a)+(2+a) = 4$
each pair has sum $4$.

Since there are $4$ distinct roots, there are $2$ pairs

the sum of all roots is \boxed{8}


[color=darkblue]Are well-known the following inequalities in a triangle $ABC\ :$

$\{\begin{array}{ccc}(-a+b+c)(a-b+c)(a+b-c) & \le & abc\\\ (b+c)(c+a)(a+b) & \ge & 8\cdot abc\end{array}\ .$ Prove that :

$\boxed{\ (b+c)(c+a)(a+b)+(-a+b+c)(a-b+c)(a+b-c)\ \ge\ 9\cdot abc\ }\ .$[/color]

[quote="Virgil Nicula"][color=darkred]Are well-known the following inequalities in a triangle $ABC\ :$

$\{\begin{array}{c}(-a+b+c)(a-b+c)(a+b-c)\ \le\ abc\\\ (b+c)(c+a)(a+b)\ \ge\ 8\cdot abc\end{array}\ .$ Prove that :

$\boxed{\ (b+c)(c+a)(a+b)+(-a+b+c)(a-b+c)(a+b-c)\ \ge\ 9\cdot abc\ }\ .$[/color][/quote]
[color=darkblue][b]Proof.[/b] The well-known relation $9\cdot IG^{2}=p^{2}+5r^{2}-16Rr$ $\implies$ $p^{2}+5r^{2}\ge 16Rr\ .$

$\{\begin{array}{c}abc=4Rrp\\\\ ab+bc+ca=p^{2}+r^{2}+4Rr\\\\ \prod (-a+b+c)=8pr^{2}\\\\ \prod (b+c)=\sum a\cdot\sum bc-abc\end{array}\|$ $\implies$

$\prod (b+c)+\prod (-a+b+c)-9abc=$

$2p\cdot (p^{2}+r^{2}+4Rr)-4Rpr+8pr^{2}-36Rpr=$ $2p\cdot (p^{2}+5r^{2}-16Rr)\ge 0\ .$[/color]

more from them :Prove that
$(b+c)(c+a)(a+b)\ge 8abc+\frac{1}{3}r^{2}(R-2r)$
:D


In triangle ABC, m_a is the lenth of median from A,
                          m_b is the lenth of median from B,
                          m_c is the lenth of median from C.

prove that:
                      a/m_a + b/m_b + c/m_c >= 2*sqrt(3).

HI amirhtlusa,

    From your hint, one may change the origin problem into a new problem relating a,b,c only(express m_a with a,b,c...), solving the new problem means solving the origin problem. However, I still could not find the direction or
anything obvious that helps the proof, what need I watch out more??

doesn't $a$ in $\frac{a}{m_a}$ stands for the lenght of the side, opposite the angle $A$?

Yes,it does, so we change  a/m_a  into  2a/sqrt(2b^2+2c^2-a^2) ??

I can prove a similar one:

       a/r_a + b/r_b +c/r_c>=2sqrt(3)


group theory
abstract algebra
superior algebra
superior algebra unsolved

Hey all

I've been stuck on this problem involving commutators:
Suppose A, B and C are all normal subgroups of some group G. Show [AB,C] = [A,B][B,C]
I tried showing directly that any element in one group is an element of the other by judicious use of the normality properties, but haven't been able to get anywhere.

Any advice would be greatly appreciated.

Use the following facts : let $ a,b,c\in G$ where $ G$ is a group
0. for subsets $ A,B$ of $ G$ then $ [A,B]\equal{}[B,A]$
1. $ [a,bc]\equal{}[a,c](c^{\minus{}1}[a,b]c)$
2.$ [ab,c]\equal{}(b^{\minus{}1}[a,c]b)[b,c]$


algebra
function
domain
limit
calculus
calculus computations

I woulk like to see  the definitions in your school or Uni textbooks , of the symbol
    $\sqrt[n] a$

 for the real number $a$.
 In Greece we introduce this symbol only for non negative numbers. In my school years $a$ could be any number if $n$ is odd.
 a) How do you use     $\sqrt[n] a$  ?
 b)  Which is the domain of $f(x) = \sqrt[3] x$ ?  For us it is $A= [0,+\infty)$ , although we know that $\mathbb{R}$
     could work very nice.

[color=red]Thanks in advance.[/color]

 Babis

What I have always used:

For $n$ odd:
$a \in \mathbb{R}$

for $n$ even:
$a \geq 0$

I think conventionally that notation is used only when $n$ is an integer larger than 2.  (The 2 is omitted for the square root.)  However, one could certainly give a meaningful interpretation to the notation for any $n \neq 0$ by setting $\sqrt[n]a = a^{1/n}$.

Dear JBL , the question is for what kind of real numbers  $a$ we define the n-nth root of $a$.For exaple , some define this root only for non negative numbers, even if n is odd. So , I'm trying to see in which other countries they have this convention. For example in USA I read that for odd n the n-nth root of $a$ has meaning for all real numbers $a$.
 But in Greece and in some other books they define it only for non negatives.
 Babis

What I can tell you is that in my books the domain is also $[0+\infty)$.

But I read in your new school book by M. Ganga(Editura GIL) , that for odd $n$ ,

    he defines the $\sqrt[n]a$ for all $a \in \mathbb{R}$.

 Which book are you refering to ?

  [u]Babis[/u]

For $n\in N^{*}$ : if $n$- odd, then $\sqrt [n]\ : \ \mathcal R\rightarrow \mathcal R$ ; if $n$- even, then $\sqrt [n]\ : \ [0,\infty )\rightarrow \mathcal R$. [b]Convention.[/b] $\sqrt [1] x=x$, for any $x\in \mathcal R$. Therefore,
$1\blacktriangleright$ If $n$- odd, then $\sqrt [n]x=y\Longleftrightarrow$ $\{\begin{array}{c}y\in\mathcal R\\\ y^{n}=x\in\mathcal R\end{array}$.
$2\blacktriangleright$ If $n$- even, then $\sqrt [n]x=y\Longleftrightarrow$ $\{\begin{array}{c}y\ge 0\\\ y^{n}=x\ge 0\end{array}$.
[b]Examples.[/b]

$\sqrt[3]{xy}=\sqrt [3]x\cdot\sqrt [3]y$ ;

$xy\ge 0\implies\sqrt [6]{xy}=\sqrt [6]{|x|}\cdot\sqrt [6]{|y|}$ ;

$\sqrt [6]{x^{2}}=\sqrt [3]{|x|}=|\sqrt [3]x|$ ;

$\sqrt [5]{x^{15}y}=x^{3}\sqrt [5]y$

$\sqrt [4]{x^{4}y}=|x|\sqrt [4]y$.

$x\in \mathcal R\implies$ $x\sqrt [3]y=\sqrt [3]{x^{3}y}$ ;

$x\sqrt y=\mathrm{sgn}(x)\cdot\sqrt{x^{2}y}$, where $x=|x|\cdot\mathrm{sgn}(x)$ a.s.o.

[b]Exercises.[/b]
$\lim_{x\to-\infty}\frac{x}{\sqrt{x^{2}+1}}=-\lim_{x\to-\infty}\sqrt\frac{x^{2}}{x^{2}+1}=-1$.

$\{\begin{array}{c}x\ge 0\ ,\ y\in\mathcal R\\\\ \sqrt x+\sqrt [3]y=6\\\\ \sqrt [6]{x^{3}y^{2}}=8\end{array}\|$ $\Longleftrightarrow$ $\{\begin{array}{c}\sqrt x+\sqrt [3]y=6\\\\ \sqrt x\cdot|\sqrt [3]{y}|=8\end{array}$ $\Longleftrightarrow$ $\{\begin{array}{c}u\ge 0\ ,\ v\in\mathcal R\\\\ u+v=6\\\\ u\cdot |v|=8\end{array}$, where $u=\sqrt x$ and $v=\sqrt [3]y$ a.s.o.

[quote="Virgil Nicula"]For $n\in N^{*}$, if $n$- even, then $\sqrt [n]\ : \ [0,\infty )\rightarrow \mathcal R$. [/quote]
Should it be: if $n$ is even $\sqrt [n]\ : \ [0,\infty )\rightarrow \ [0,\infty )$

[quote="Virgil Nicula"]Emiraga, why surjective ?![/quote]It's difficult to say why for a [u]definition[/u]. Usually elementary [i]real[/i] functions are defined as single-valued. You said it yourself:
[quote="Virgil Nicula"]If $n$- even, then $\sqrt [n]x=y\Longleftrightarrow$ $\{\begin{array}{c}y\ge 0\\\ y^{n}=x\ge 0\end{array}$.[/quote]
$y\ge 0$

Is there any difference between $\sqrt[n] x$ and $x^{\frac{1}{n}}$? (because rational powers are defined only for positive numbers, I had this question in a contest )

[b]YES![/b]

 It is a different notation. For possitive numbers it is the same. For negatives we have problem. See for example this :

  $-1 = \sqrt[3]{-1}= (-1)^{\frac{1}{3}}= (-1)^{\frac{2}{6}}= \sqrt[6]{(-1)^{2}}= \sqrt[6]1 = 1$ !

 For this reason we define the rational (fraction) powers only on possitive or nonnegative numbers.
 But I want to see the  many opinions on this subject.

[u] Babis  [/u]


function
ratio
trigonometry
complex numbers
complex analysis
complex analysis unsolved

I believe you have to use the Weiestrass [b]M-Test[/b] for these.

Show that each of the following is an entire funtion of z:

A)
inf        z^(n)
Sum    ------------
n=1    (n!)^(1/2)

B)
inf        z^(n)
Sum    --------------
n=1     2^(n)^(2)

C)
inf         1          1
Sum    --------  ---------
n=1    2^(n)      n^z)

D)
inf        sin(nz)
Sum    ------------
n=1        n!

Thanks in advanced.  Maybe the ratio test needs to be used too.

Well if you read what the weirstrass M-test is,

http://en.wikipedia.org/wiki/Weierstrass_M-test

The first one you just need to prove that

$ \sum_{n\equal{}1}^\infty \frac{|z|^n}{\sqrt{n!}}$ converges. This is obvious by the ratio test.

Then the original converges uniformly for any $ z$.

Then presumably, since this function has a power series that converges everywhere, it is analytic.

B) is basically the same.

For part C), you need to know what $ |n^z|$ is... If $ z\equal{}re^i\theta$, then $ |n^z|\equal{}n^r$

Then the rest is obvious.

D), Consider the series that is

$ \frac{\cos nz\plus{}i\sin nz}{n!}$ which is equal to $ \frac{(e^z)^n}{n!}$ which is just $ e^{e^z}$ so that series converges so of course the real part converges. Then done as before.


geometry
calculus
integration
vector
number theory
Ring Theory
superior algebra

Let $ k\subseteq k'$ be fields and $ k'$ normal as a field extension. Let $ G\equal{}Aut_k(k')$ and $ R$ an affine ring over $ k$ and let $ R'\equal{}R\otimes_k k'$. Show that the map $ Spec(R')\rightarrow Spec(R)$ induces a bijection between $ Spec(R)$ and the $ G$-orbits of $ Spec R'$

The proof uses the intermediate fixed field $ P \equal{} (k')^G$ for a two-step approach: $ k'|P$ is a Galois extension and $ P|k$ is purely inseparable (French: extension radicielle). The latter implies that $ Spec(P)\to Spec(k)$ is universally injective, in particular $ Spec(R\otimes_k P)\hookrightarrow Spec (R)$ is injective (*).

What I want to show is that $ Spec(R\otimes_kk')/G\overset{\varphi_1}{\to} Spec((R\otimes_kk')^G)\overset{\varphi_2}{\to}Spec(R\otimes_kP) \overset{\varphi_3}{\to} Spec(R)$ is a composition of bijections.

A collection of facts proves the claim:

1. Since $ k'|k$ is normal, in particular finite, hence integral, thus $ Spec(k')\to Spec(k)$ is (universally) surjective. In particular $ Spec(R\otimes_kk')\to Spec(R)$ is surjective. This implies that all maps involved are surjective. Together with (*) we get that $ \varphi_3$ is bijective.

2. $ P$ is defined by the exact sequence of $ k$-vector spaces
\[ 0\to (k')^G\to k'\rightrightarrows \bigoplus_{\sigma\in G}k'\]
where the upper arrow is defined by $ a\mapsto (a)_{\sigma\in G}$, the lower by $ a\mapsto (\sigma(a))_{\sigma\in G}$. Tensoring with the $ k$-vector space $ R$ yields the exact sequence defining $ (R\otimes_kk')^G$
\[ 0\to R\otimes_k(k')^G\to R\otimes_kk'\rightrightarrows \bigoplus_{\sigma\in G}R\otimes_kk'\,.\]
Thus we can read off  $ (R\otimes_kk')^G \equal{} R\otimes_k(k')^G \equal{} R\otimes_kP$, hence $ \varphi_2$ is a bijection.

3. To see that $ \varphi_1$ is injective simply mimic the well known proof in number theory that prime ideals in the Galois extension lying above the same prime ideal are conjugated by a Galois homomorphism. The proof makes heavy use of the Chinese remainder theorem.

Note that
- it is true that all maps are homeomorphims which is a bit more than claimed. But is follows easily from that the morphisms are integral.
- $ P \equal{} k$ in characteristic 0.


Solve the system of equations: \[ \left\{\begin{array}{lll} 2y(x^2\minus{}y^2)&\equal{}&3x\\ x(x^2\plus{}y^2)&\equal{}&10y\end{array}\right.\]

if we multiply equations we find $ x^4 \minus{} y^4 \equal{} 15$ one of x and y are x= 2 and y=1.

Missing some solutions.  Is there division by zero? :wink:

Furthermore, the problem doesn't specify a restriction to integers.

Well, if there isn't a restriction for only rational numbers, then shouldn't there be infinite amount of solutions?

[url=http://en.wikipedia.org/wiki/B%C3%A9zout%27s_theorem]Bezout's theorem[/url] tells us that there are at most $ 9$ solutions.  Note that this is equivalent to solving for one variable in terms of the other and plugging in to produce a polynomial of degree $ 9$ (although this construction is not always explicitly possible).

Edit:  [hide="Solution"] Let $ a \equal{} \frac{y}{x}, b \equal{} x^2$.  This gives

$ 2ab(1 \minus{} a^2) \equal{} 3$
$ b(1 \plus{} a^2) \equal{} 10a \implies b \equal{} \frac{10a}{1 \plus{} a^2}$.

Substituting into the other equation, $ 20a^2(1 \minus{} a^2) \equal{} 3(1 \plus{} a^2)$.  Letting $ c \equal{} a^2$, we have

$ 20c \minus{} 20c^2 \equal{} 3 \plus{} 3c \implies$
$ 20c^2 \minus{} 17c \plus{} 3 \equal{} 0 \implies$
$ (5c \minus{} 3)(4c \minus{} 1) \equal{} 0 \implies$
$ c \equal{} \frac{3}{5}, \frac{1}{4}$

from which the other solutions can be calculated.  [/hide]

For $ y\neq 0$, dividing both sides gives $ \frac {2y(x^2 \minus{} y^2)}{x(x^2 \plus{} y^2)} \equal{} \frac {3}{10}\cdot \frac {x}{y}$

$ \Longleftrightarrow 2\cdot \frac {y}{x}\cdot \frac {\left(\frac {x}{y}\right)^2 \minus{} 1}{\left(\frac {x}{y}\right)^2 \plus{} 1} \equal{} \frac {3}{10}\cdot \frac {x}{y}$

Let $ z \equal{} \frac {x}{y}$, we have $ 3z^4 \minus{} 17z^2 \plus{} 20 \equal{} 0$, yielding $ z \equal{} \pm 2,\ \pm \sqrt {\frac {5}{3}}\Longleftrightarrow \frac {y}{x} \equal{} \pm 2,\ \pm \sqrt {\frac {5}{3}}$.

Needless to say, $ (x,\ y) \equal{} (0,\ 0)$ is one of the solutions.

Are you guys sure? I multiply both to get $ x^4 \minus{} y^4 \equal{} 15$, so $ x \equal{} \pm \sqrt [4]{y^4 \plus{} 15}$. Also by the first equation we have $ 2yx^2 \minus{} 3x \minus{} 2y^3 \equal{} 0$, so by the quad. formula, we have $ x \equal{} (3 \pm \sqrt {16y^4 \plus{} 9})/4y$. Setting both equal and simplifying gives $ 256y^4(y^4 \plus{} 15) \equal{} (3 \pm \sqrt {16y^4 \plus{} 9})^4$. I just saw that $ y^4 \equal{} 1$ is a soultion. Then, x would equal 2.

[quote="dgreenb801"]Are you guys sure? I multiply both to get $ x^4 \minus{} y^4 \equal{} 15$, so $ x \equal{} (y^4 \plus{} 15)^(1/4)$.[/quote]

Is limited $ x>0$?

Oh, nevermind, what I wrote matches what kunny said but doesn't include everything.

[quote="Altheman"]$ 20y^2(x^2 \minus{} y^2) \minus{} 3x^2(x^2 \plus{} y^2) \equal{} (3x^2 \minus{} 5y^2)(4y^2 \minus{} x^2) \equal{} 0$[/quote]

Oh! Best soution. :lol:


calculus
integration
quadratics
function
trigonometry
calculus computations

For any quadratic functions $ f(x)$ such that $ f'(2)\equal{}1$, evaluate $ \int_{2\minus{}\pi}^{2\plus{}\pi}f(x)\sin\left(\frac{x}{2}\minus{}1\right) dx$.

[hide="Straightforward."]This looks better with a change of variables. Let $ u\equal{}x\minus{}2$ and $ g(u)\equal{}f(u\plus{}2)$ so that $ g(x\minus{}2)\equal{}f(x)$ and $ g'(0)\equal{}1.$

$ \int_{2\minus{}\pi}^{2\plus{}\pi}f(x)\sin\left(\frac{x}{2}\minus{}1\right)\,dx\equal{}\int_{\minus{}\pi}^{\pi}g(u)\sin\left(\frac{u}{2}\right)\,du$

$ \equal{}\int_{\minus{}\pi}^{\pi}(a\plus{}u\plus{}cu^{2})\sin\left(\frac{u}{2}\right)\,du$

But $ \int_{\minus{}\pi}^{\pi}(a\plus{}cu^{2})\sin\left(\frac{u}{2}\right)\,du\equal{}0$ because the integrand is odd.

That leaves $ \int_{\minus{}\pi}^{\pi}u\sin\left(\frac{u}{2}\right)\,du\equal{} 2\int_{0}^{\pi}u\sin\left(\frac{u}{2}\right)\,du$

$ \equal{}\left.\minus{}4u\cos\left(\frac{u}{2}\right)\right|_{0}^{\pi}\plus{}4\int_{0}^{\pi}\cos\left(\frac{u}{2}\right)\,du$

$ \equal{}\left.8\sin\left(\frac{u}{2}\right)\right|_{0}^{\pi}\equal{}8.$[/hide]


Consider the function $ f: R\minus{}>R$ $ f(x)\equal{}\frac{1\minus{}2x}{x\plus{}2}$.Solve the equation $ (f\circ f\circ...\circ f)(x)\equal{}\minus{}1$($ n$ times $ f$), where $ (f\circ f)(x)\equal{}f(f(x))$

First list the complete cycle of $ f(x)$.

$ f(x) \equal{} \frac {1 \minus{} 2x}{x \plus{} 2}$

$ f^2 (x) \equal{} x$

So we find that

If $ n \equal{} 2k \plus{} 1$, $ x \equal{} 3$.

If $ n \equal{} 2k$, $ x \equal{} \minus{}1$.


I was wondering if anyone has used any of the Add-On Curriculum modules for GSP?  They seem to be very interesting but I would like to hear what others thought of them.

[quote="joml88"]Add-On Curriculum modules for GSP[/quote]
Can you please tell me what it is?

It's a geometry program (makes geometry drawings on the computer).

I have the Geometer's Sketchpad, but what is Add-On Curriculum modules?


No, the tens digit is the last but one digit, the one coming just before the units digit.

[quote="Altheman"]$0>n>10$[/quote]
Small correction: it should be $0<n<10$

It can be written in the form:-
$ 3^{m}\equal{}10 \times x\plus{}1,3,7,9$
It can easily be shown that if it is 1, then it is 1 mod 4, if it is 3 then it is 3 mod 4, if it is 7 it is 3 mod 4  and if it is 9 then it is 1 mod 4.
(We know that $ 10x\plus{}3\equal{}3^1,3^5,3^9....$
$ 3^{5}\equiv 3\mod{4}$
so,
$ 3^{9}\equiv 3\mod{4}$
,so on and similarly the others.

Taking each of the four cases , we see that $ x\equiv 0\mod{2}$

Yay, my first proper post :D .

Observe that 3^3=27...
Now after this unit's place of 3^n can be 1,3,9,7,....the cycle repeats, observe each leaves a rem of 0 or 2, which when added to 2 makes it even, thus proved.

[quote=madhusudan kale][quote="Singular"]We are going to look at 3^k mod 20.  It repeats 3,9,7,1.  Hence result.[/quote]

Could you please elaborate.[/quote]

I think Singular is referring to the fact that when each of those numbers is multiplied by 3, their tens digit is a 2. Hence the sum of the tens digits must be even is what he is arguing we can infer from that....not entirely sure if that's a rigorous argument.

[quote=Altheman][hide="more completely..."]

we know that if $3^m \mod{20} \equiv n$ where $0>n>10$, then it has an even tens digit because the remainder will not effect the tens digit if it is less than 10, and if it is in mod 20, then any number times 20 has an even tens digit

note that 3 and 20 are relatively prime, so using euler's extention of FLT, (note that $\phi(20)=8$)
so $3^{8k+m} \mod{20} \equiv 3^m \mod{20}$

so we only need to go through m=0,1,2,3,4,5,6, and 7, and show that the remainders are all less than 10, then all the other powers are also proven

$3^{0} \mod{20} \equiv 1$
$3^{1} \mod{20} \equiv 3$
$3^{2} \mod{20} \equiv 9$
$3^{3} \mod{20} \equiv 7$
$3^{4} \mod{20} \equiv 1$
$3^{5} \mod{20} \equiv 3$
$3^{6} \mod{20} \equiv 9$
$3^{7} \mod{20} \equiv 7$

thus every remainder $\mod{20}$ for a power of $3^n$ is 1,3,7, or 9
so the tens digit must be even because any number mod 20 where the remainder is less than 10 will have an even tens digit

[/hide][/quote]

what is euler's extension of FLT

[quote=RMO 1993 P2]Prove that the ten's digit of any power of 3 is even.[/quote]
[b][color=#000]Solution:[/color][/b] Suppose that the power of $3$ is $3^k$. We'll do a quick induction on $k$. First check that $3^3=27$, $3^4=81$, $3^5=243$, $3^6=729$ and $3^7=2187$

Suppose, for some $m \in \mathbb{N}$, $3^m$ has an even ten's digit, say $x$. So:
1) If the unit's digit of $3^m$ is 1, then ten's digit of $3^{m+1}$ $\implies$ $3x \equiv \ell \pmod{10}$
2) If the unit's digit of $3^m$ is 3, then ten's digit of $3^{m+1}$ $\implies$ $3x \equiv \ell \pmod{10}$
3) If the unit's digit of $3^m$ is 7, then ten's digit of $3^{m+1}$ $\implies$ $3x+2 \equiv \ell \pmod{10}$
4) If the unit's digit of $3^m$ is 9, then ten's digit of $3^{m+1}$ $\implies$ $3x +2\equiv \ell \pmod{10}$
(Where, $\ell \in \{0,2,4,6,8\}$). Hence, our Induction is complete and all powers of $3$ have an even ten's digit! $\qquad \blacksquare$

Really sorry to bump this trivial problem but I thought this approach was nice.$\pmod{20}\Rightarrow$ there exists $t$ such that $20t<3^n<20t+10$ so $2t<\frac{3^n}{10}<2t+1$ hence $
 [\frac{3^n}{10}]=2t$ ($[.]$ is the floor function).Note that $[\frac{3^n}{10}]$ is the number formed by removing the last digit of $3^n$ and moreover this is even.The result follows

Whoops, I fell asleep while solving this. Here's the solution:

Notice that the last digit of every number in the form $3^k$ (where $k$ is a non-negative integer) goes in the pattern: $3, 9, 7, 1$.

The last digit will be $3$ when $k \equiv 1\pmod{4}$, the last digit will be $9$ when $k \equiv 2\pmod{4}$, the last digit will be $7$ when $k \equiv 3\pmod{4}$ and the last digit will be $1$ when $k \equiv 0\pmod{4}$. We can separate this problem into 4 different cases.

Case 1: The last digit of $3^k$ is $1$ - This means that $k \equiv 0\pmod{4}$. $k$ can be written as $4n$ where $n$ is a non-negative integer. So, $3^k = 3^{4n} = 81^n$. We want the ten's digit to be even, so:
$$\frac{81^n - 1}{10} \equiv 0\pmod{2}$$
$$\implies 81^n \equiv 1\pmod{20}$$
This is clearly true as $81 \equiv 1\pmod{20}$. So we have proved our statement for the first case.

Case 2: The last digit of $3^k$ is $3$ - This means that $k \equiv 1\pmod{4}$. $k$ can be written as $4n + 1$ and we can proceed to do the same thing as Case 1 to get that $81^n\cdot3^1 \equiv 3\pmod{20}$ which is true.

Case 3: The last digit of $3^k$ is $9$ - This means that $k \equiv 2 \pmod{4}$. $k$ can be written as $4n + 2$ and we can proceed to do the same thing as Case 1 to get that $81^n\cdot3^2 \equiv 9\pmod{20}$ which is also true.

Case 4: The last digit of $3^k$ is $7$ - This means that $k \equiv 3\pmod{4}$. $k$ can be written as $4n + 3$ and we can proceed to do the same thing as Case 1 to get that $81^n\cdot3^3 \equiv 7\pmod{20}$ which is true as $81^n \equiv 1\pmod{20}$ and $3^3 \equiv 7\pmod{20}$.

We are done $\blacksquare$


HERE GOES THE PROBLEM
Given are 3 nonnegative integers: k, l and m. In one transformation one considers two of the three numbers, say WLOG k and l, and replaces them by k+l and |k−l|. Find out if there it is possible to obtain the triplet (0, 0, 0)by making such transformations.

Do I miss something or the problem is trivial?

It is clear that if you start with (k,l,n) where k,l,n are nonnegative integers, the only way to obtain (0,0,0) is to start with (0,0,0) since :
If at some step we choose k and l with k  \geq  l, then we replace them with k' = k+l and l' = k-l, which are nonnegative too and such that k'+l'+n' = k+l + k-l + n = 2k +n  \geq k + l + n.
Then the operation does not decrease the sum of the numbers. So if we start from k+l+n > 0 we will never reach (0,0,0).

Pierre.

This reminds me of following problem:

Given are 3 non-negative integers: (a,b,c)
One transformation is to reduce the value of one integer x and double the size of integer y by replacing x with x-y and y with 2y. This is only allowed, when x >= y.
Is it possible to make one field zero (reaching (x,y,0), the order of the integers doesn't matter)?

This shouldn't be so easy  :D

i actually had solved the problem the same way as pierre but didnt think it would be that simple :D .that's why i made the post

Let's solve the Daniel's problem by the affirmative :
Let a,b,c be the initial number. Wlog, we may assume that a \leq b \leq c, and that a > 0 if not the problem is trivial.

We will prove that after a finite number of steps we may reach a configuration (a',b',c') with min(a',b',c')  \leq  a-1. Thus, repeating this process we will eventually reach (x,y,0) for some nonegative integers x,y after a finite number of steps.

Let k be such that b = ka + p with 0  \leq p < a, and k = k_0 + 2*k_1 + ... + 2 n *k_n be the binary expansion of k.
Then b = p + sum a*k_i*2 i  and c  \geq  p + sum a*k_i*2 i.

Let r be the least subscript such that k_r = 1. We use r times the operation (c,a) - > (c-a, 2a) to reach (2 r *a, b, c - 2 r *a + a).
Then we use (b, 2 r *a) - > (b - 2 r *a, 2 r+1 *a), and reach (2 r+1 *a, p + sum{i > r} 2 i *a*k_i , c - 2 r *a + a).
If r = n, we have p + sum{i > r} 2 i *a*k_i = p < a, and we are done.
If r < n, then let r_2 be the least subscript > r such that k_(r_2) = 1, and we do it again as above....and so on, until we reach (a*2 n , a*2 n  + p, c') and then we obtain (a*2 n+1 , p, c'), and we are done.

Pierre.

This problem from poland TST 2002 was a bit diffrent: to obtain the triplet (a, 0, 0) by making such transformations, a is nonnegative integer.

The immediate answer is that [b]not always [/b]can we do that. If we start with $0<x<y$ we cannot reach $0,0,a$. A move involving $0$ does nothing, so we must consider $x<y\mapsto y-x< y+x\mapsto 2x < 2y \mapsto \cdots$, never reaching $0,0,a$.

Most likely the challenge of the problem was to characterize those starting triplets from which we [b]can[/b] reach  $0,0,a$.

A move involving $0$ do something, for example we can replace $ (1,0) $ with $ (1,1) $, because $1+0=1$ and $1-0=1$.

True - for some reason, a hasty conclusion of mine :)

Let the numbers be $k,l,m$ at beggining.Their sum is: $k+l+m$.$WLOG$ suposse that we do our first move on $k$ and $l$ and also $WLOG$ $k>=l$.After the move the numbers look like : $k-l,k+l,m$.Their sum is:
$k-l+k+l+m=2k+l>=k+l+m$.This way we see that sum can only raise after every move so we can never reach $(0,0,0)$.

it is not written here that 3 must come to zero.
Some were asked if the two could come to zero.
This is a very wrong idea.

The condition of the issue is also different with Polish MO finals 2002 problem 3 it is said whether there can ever be two 0's

If this is the case, then the exact sum always increases and never reaches 3 zeros


I recently received the invitation to seminar, and from what i've read in the invitation letter, it sounds really fun and nice :) .
However, there is a little problem:
[QUOTE]RULES:5.Appropriate dress(eg jackets and ties for young men( :rotfl: )) is required for the banquets.[end quote]
So, this means there's some kind of dress code for banquets. I want you ppl who have gone to seminar before please write what i am supposed to wear???, cause i don't want to look like a total idiot during the banquet. I just ask this question that may seem stupid to some of you because I haven't attended any such banquets in my life.
And btw, if others want ot ask the questions about the seminar(such as transportation arrangements, etc) or write any comments, feel free to post in this trend.

You already got an invitation? Wow, that's fast.
You can take a look at Yufei's photo to find out what to wear.
[url]http://ca.pg.photos.yahoo.com/ph/z_yufei/my_photos[/url]
___________________________________________________________________________
EDITED: A grammatic error fixed. :blush:  I looked at some photos. I should bring a blindfold this time.

Thanks a lot. now i know. :)
EDIT: Now it not only sounds like fun, it looks like fun, too. :lol:

Seminar is THE BEST.  End of discussion.  You'll love it.  I might even show up on a few of the nights (considering I'll be at school all day anyway, a walk over to renison after classes might not be a bad idea)

BTW.  If you make the IMO team, you have to start a post on here somewhere with 3 words that begin with the letters I, M, O, respectively (since you're not allowed to reveal the team yet).  For example: "It may only be a matter of time before seminar starts, but I can't wait!"

Get it?

BWAHAHAHAHAH!

Oh yeah, and for the appropriate dress thing... be sure to wear sneakers like peng did.

[quote="lightrhee"][quote="Elyot"]
BTW.  If you make the IMO team, you have to start a post on here somewhere with 3 words that begin with the letters I, M, O, respectively (since you're not allowed to reveal the team yet).  For example: "It may only be a matter of time before seminar starts, but I can't wait!"

Get it?

BWAHAHAHAHAH!
[/quote]
[b]I[/b]n [b]m[/b]y [b]o[/b]pinion, rem will make the team without problem :D[/quote]

geez... you couldn't be a little more subtle than that?  I mean... you had to quote me AND use bold letters?

Meh, we all knew you'd make the team anyway.

[quote="Elyot"]
Oh yeah, and for the appropriate dress thing... be sure to wear sneakers like peng did.[/quote]

Hahahahaha. I laughed so hard at this :D

Can you get invited to seminar with 144 on Fermat?

I think me and Lee are the only ones from wci to get invited. That's what Burns told me. But if you get in or don't get in totally depends on how others do compared to you. There are really a lot of people who got 144 this year.

I got in with less that 144 last year, but then again, I'm a girl... they make it easier for us :P

They tend to invite provincial champions of each of the contests, provided you get above a certain cut-off, so there's a good geographical representation there.  They seem to bias it toward girls for sure.  So your best chance of getting there would be to like... be a female in PEI or something.  Sadly enough, there were people there with 120-something on Fermat.  Really though, I don't know many people who deserved an invite and didn't get one.

Ah, wow, its nearing seminar time again!  Bleh, too bad I'll only get in if I did decently on Euclid.

Hehee, yes, its jacket and tie.  Or no jacket if you don't want to bring one.

[quote="Sunny"]Ah, wow, its nearing seminar time again!  Bleh, too bad I'll only get in if I did decently on Euclid.  [/quote]

I'm hoping you'll get lucky enough to be invited though.  They're kinder to girls when they select who gets to come to seminar.

___________________________________
EDITED: Oops! I was going to quote myself, and I edited this thing!! So, the whole thing got deleted. :(

[quote="Elyot"]The seminar news that more of you will REALLY care about!

- Sportsworld is GONE!  Kaput!  Vamoosed!  Essentially, they went bankrupt and sold off everything at a huge auction.  So I think the annual seminar trip to Sportsworld will have to be replaced with something else!

- Waterloo's DDR Machine is GONE!  Yep, the upper cove management decided to move it to Toronto this summer, citing reasons of "it's more profitable there".

As a solution to both of the above problems, I propose that the annual trip to Sportsworld be replaced with a party at Elyot's (yes, I have DDR!)

Hahahah... just kidding, of course.  I don't want to get too many people in trouble this year!  Maybe instead I can make a visit and teach you all how to play The Settlers of Catan.

It might not be a bad idea to start a seminar roll call to see how many AOPS/Mathlink people make it this year![/quote]
 :o
Can this get any worse?
So we will have to do Math all the time during the seminar. :)

[quote="Elyot"]The seminar news that more of you will REALLY care about!

- Sportsworld is GONE!  Kaput!  Vamoosed!  Essentially, they went bankrupt and sold off everything at a huge auction.  So I think the annual seminar trip to Sportsworld will have to be replaced with something else!

- Waterloo's DDR Machine is GONE!  Yep, the upper cove management decided to move it to Toronto this summer, citing reasons of "it's more profitable there".

As a solution to both of the above problems, I propose that the annual trip to Sportsworld be replaced with a party at Elyot's (yes, I have DDR!)

Hahahah... just kidding, of course.  I don't want to get too many people in trouble this year!  Maybe instead I can make a visit and teach you all how to play The Settlers of Catan.

It might not be a bad idea to start a seminar roll call to see how many AOPS/Mathlink people make it this year![/quote]

WHAT?!  NO!!! But but but.....DDR....Sportsworld....SEMINAR TRADITION!!!!!!!!!!!!!!!!  *sniff*

Just won't be the same.....

[quote="lightrhee"]
Elyot(or anyone else who knows), do you know when the seminar starts and ends?
[/quote]
I need this information as fast as I can(especially the latter). Please answer if you know when it ends. :omighty:

Ahhhh well **** that then.  SATs on the 17th, and Summer Program starting sometime the week of the Seminar.  That's what I get for compromising with ARML  ;) .

Have fun guys.  Eat cheese.  :P

[quote="rem"]June 11 - June 17.[/quote]
Thanks :thumbup:


induction
pigeonhole principle
strong induction

Hello,

I struggled on the following problem for dozens of hours and still can't find the answer!! Please help me before I go crazy!!

Here's the problem:
Given a set of n + 1 numbers out of the first 2n natural numbers 1,2...,2n, prove by induction that there are two numbers in the set, one of which divides the other.

It might be a strong induction proof!

It's not induction; it's pigeonhole.

A hint: the quotient of some pair will be a power of 2.

[hide="Pigeonhole Solution"]

Let $f(n)$ be the greatest odd factor of $n$.  Observing that $f(2n)=f(n)$, there are exactly $n$ different values of $f(n)$ over the integers from $1$ to $2n$.  Then, if we choose $n+1$ of these numbers, by the Pigeonhole Principle, we must have two distinct numbers $a<b$ such that $f(a)=f(b)$.  Then $b=2^{k}\times a$ for some positive integer $k$, and $a|b$.[/hide]

anouther solution by pigeonhole :
We will try to build some "boxes" in this case some chains :
C1={1,2,4,8,....}
C2={3,6,12,.....}

Cn={2n-1}

and now we will use pigeonhole : there are n boxes and n+1 objects and the conclusion is easy to see......

Remark: this is a problem of Erdos[/quote]

Thank you all!

One more thing, why in TomciO's proof, there exist that so called number d, and what is it?

[quote="TomciO"]But it can be proved with induction also.

Suppose that the thesis works for $2n$. We will prove that it works for $2n+2$. So, suppose that there exist a set $A$ of $n+2$ numbers (from the $\{1, 2,..., 2n+2\}$) such that our claim is not true. If there were at least $n+1$ numbers from the set $\{1, 2, ..., 2n\}$ in $A$ then we would be done, because of the induction hypothesis. So, $2n+1, 2n+2 \in A$. Now, it's clear that $n+1 \not \in A$ since $n+1|2n+2$. But, because of the induction hypothesis, we know that if we would add $n+1$ to the set $A$ the claim would be true, so there exist number $d$ such that $d|n+1$ or $n+1|d$. The latter is not possible since $d \not = n+1$ and $d \leq 2n < 2(n+1)$. So $d|n+1$. But then $d|n+1|2n+2$, a contradiction.[/quote]

I will just clarify the parts I don't understand a bit.

What do you mean by adding the number $n+1$ to the set $A$ ? Does that make $A$ a set of $n+3$ numbers? Or is it just considering the the number $n+1 \in A$ and $A$ still being a set $n+2$. If so, are you swapping the number $n+1$ with the number $d$?

Also, what makes you say that $d|n+1$ or $n+1|d$ ? The claim states that at least one number is to divide another. But what makes you say that $n+1$ is one of those two numbers. $n+1$ could be a prime number after all, or simply a number indivisible by any numbers < $n+1$ contained in the set A.

There are supposed to be $n+2$ elements in $A$. $2$ elements are $2n+1,2n+2$. Thus, there are $n$ elements in the original set (and amongst these elements, the desired may not be true). So if you add the element $n+1$ to the set, then there would be $n+1$ elements in the original set and by the induction hypothesis, the desired is true. Thus, if you add $n+1$, there exists a $d$ that either divides $n+1$ or $n+1$ divides $d$. We are just considering what happens when we only consider the elements in the original set.

If $n+1$ is prime, then $d=1$.


Dandu-se numerele reale pozitive $\ a,b,c,d$ si numarul natural $\ n$
gasiti cele mai bune constante $\epsilon,\alpha,\gamma$, astfel incat sa avem adevarata inegalitatea
$\ ((a-1)(b-1))^{2n+1}+((c-1)(d-1))^{2n+1}\geq$
$\geq\epsilon+\alpha n(ab+cd)-\gamma (n+1)(a+b+c+d)+ab+cd$

Kosmine, asta tot asa nu prea atrage?? :blush:

Mie asa in timpul liber nu prea-mi plac inegalitatile astea cu 141042141 d variabile...si constante si etc :|

Asa in timpul liber imi plac dastea mijto..gen..suma din ceva >= n sau produs din ceva >= n...sau ceva trigonometrie <= sau >= cu n

:) :blush:

Si de accea postez rezolvari numai la astfel de inegalitati :)

Bine mai...asteapta ca o sa propun si d-alea de care zici.Intr-adevar, asta e mult prea usoara si cam lunga, poate d-aia nu vrei sa postezi. ;)

nu ma...dar nu-mi place :D poate sa fie ea usoara sau grea...problemele de genu asta le fac doar cand e nevoie neaparat :D acu la 2:37 a.m neee :)


[quote="ch1n353ch3s54a1l"][quote="boxedexe"][b]Problem.[/b] Find the number of ordered triples of sets $(A,B,C)$ such that $A\cup B\cup C=\{ 1,2,\hdots, 2003\}$ and $A\cap B\cap C=\emptyset .$[/quote]
[hide="...Er..."]
By the second condition they all share an element, and there are 2003 possibilities for that... then every other element in $\{1,2,\ldots , 2003\}$ have to be put in any of $A,B,C$. For each element there are 7 possible ways: $A,B,C,AB,BC,AC,ABC$ and there are 2002 of those...$2003\cdot 7^{2002}$... :huh: [/hide][/quote]

Am I missing something? If they all share an element then the intersection would not be an empty set...  :maybe:

oh oops, I thought it said $A\cap B\cap C\neq \emptyset$ :huh: wonder why...
[hide="Tried getting somewhere..."]
So then they can't share any elements... there are $2003!$ ways to arrange the numbers in the set and the number of ways to choose how many elements in A,B,C are the number of non-negative solutions to $A+B+C=2003$ which is $\binom{2005}{2}$...Since each permutation of the 2003 elements is counted and you can count forward from 1 in the set to the value of A in $A+B+C=2003$ , then to the value of B, then the rest is C...but then we've over counted because the elements might be permuted within A,B,C....argh[/hide]

How about just this? It's basically the same thing you said earlier...

[hide]

Each number must be in atleast one set, but not in all 3. So there are 6 combinations of what it can be in, A, B, C, AB, AC, or BC. So then the answer is $6^{2003}$

[/hide]

[quote="tjhance"]How about just this? It's basically the same thing you said earlier...

[hide]

Each number must be in atleast one set, but not in all 3. So there are 6 combinations of what it can be in, A, B, C, AB, AC, or BC. So then the answer is $2003^{6}$

[/hide][/quote]

Oh I see...thanks

Not $6^{2003}$?  :huh:


trigonometry
geometry proposed
geometry
U sviti matematyky

Points $ A, B, C$ and $ D$ are chosen on a circle in such a way that the point $ S$ of the intersection of the lines $ AB$ and $ CD$ lies outside the circle and the point $ T$ of intersection of the lines $ AC$ and $ BD$ lies inside the circle. Let points $ M$ and $ N$ lie on chords $ BD$ and $ AC$ respectively and $ K$ denotes point of intersection of the lines $ ST$ and $ MN$. Prove that $ \frac{MK}{KN}\equal{}\frac{TM}{TN}.\frac{BD}{AC}$.

I will assume the points $ M,N,T$ to be distinct (otherwise: guess what will happen :P)

It is clear that $ \angle SBT = \angle SCT$. Also, the triangles $ SBD$ and $ SCA$ are similar, so
\[ \frac {SC}{SB} = \frac {AC}{BD}.
\]
Also, using the law of sine on the triangles $ TMK$ and $ TNK$ respectively, we obtain
\[ \frac {TM}{TN} \cdot\frac {KN}{KM} = \frac {\sin\angle NTK}{\sin\angle MTK}.
\]
Thus,
\begin{align*} \frac {TM}{TN}\cdot\frac {KN}{KM} & = \frac {\sin\angle NTK}{\sin\angle MTK} \\
& = \frac {\sin\angle CTS}{\sin\angle BTS} \\
& = \frac {\frac {SC}{ST}\sin\angle SCT}{\frac {SB}{ST}\sin\angle SBT} = \frac {SC}{SB} = \frac {AC}{BD} \end{align*}
$ \Box$


Let $ABCD$ is an inscribed quadrilateral.
$K\in{AC\cap{BD}}$,in the triangle $AKD$ there is a point $P$ such that $\angle{APC}=\angle{ADC}+90^0$ and $\angle{BPD}=\angle{BAD}+90^0$.
Prove that the diogonals of the quadrilateral formed by projections of $P$ on the side of $ABCD$ are prpendicular.

Posted before as [url=http://www.mathlinks.ro/Forum/topic-80597.html]Nice problem[/url].


(i) Consider two positive integers $a$ and $b$ which are such that $a^a b^b$ is divisible by $2000$. What is the least possible value of $ab$?
(ii) Consider two positive integers $a$ and $b$ which are such that $a^b b^a$ is divisible by $2000$. What is the least possible value of $ab$?

$2000 = 2^4 \cdot 5^3$

Introducing other prime factors would only increasing product uselessly.

$10$ is an obvious minimum in the first case which is reached with $1^1 \cdot 10^10$

In the second case, wlog, $2 | a$ and $5 | b$. Minimum is for $4^5 \cdot 5^4$

Isn't the answer for problem 1
$a=10, b=1$?

Yeah, the answers are 10 and 20 respectively..


linear algebra
matrix
abstract algebra
Ring Theory
linear algebra unsolved

Find the number of homomorphisms of the rings $\mathrm{Mat}_{2\times2}(\mathbb{C})\to\mathrm{Mat}_{3\times3}(\mathbb{C})$ such that the image of $2\times2$ identity matrix is $3\times3$ identity matrix.
(V. Mazorchuk)

The general result is this:

$(\#)$ Let $k$ be a field, and let $m,n$ be positive integers. There is a ring homomorphism $\varphi: \mathcal M_{n}(k)\to\mathcal M_{m}(k)$ iff $n|m$.

(here, all rings and algebras have a unit, and all ring or algebra homomorphisms are assumed to preserve units)

The version of $(\#)$ with "ring homomorphism" replaced by "$k$-algebra homomorphism" has been discussed before on the forum (or at least the more difficult implication has :)). See, for example, the last message in [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=algebra&t=4923]this[/url] topic, and also [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=module+simple&t=90862]this[/url]. The stronger statement $(\#)$ follows if we prove that whenever there is a ring homomorphism $\varphi: \mathcal M_{n}(k)\to\mathcal M_{m}(k)$, we can also define a $k$-algebra homomorphism $\tilde\varphi: \mathcal M_{n}(k)\to\mathcal M_{m}(k)$:

For $i,j=\overline{1,n}$, let $e_{ij}$ be the $n\times n$ matrix having $1$ in position $(i,j)$ and $0$ everywhere else. The elements $e_{ij}$ of $\mathcal M_{n}(k)$ form a $k$-basis, and satisfy the relations $e_{ij}e_{kl}=\delta_{jk}e_{il}\ (*)$ ($\delta_{jk}$ being the Kronecker delta). If $f_{ij}=\varphi(e_{ij})$, then the elements $f_{ij}\in\mathcal M_{m}(k)$ also satisfy $(*)$. Now simply define $\tilde\varphi\left(\sum\alpha_{ij}e_{ij}\right)$ to be $\sum\alpha_{ij}f_{ij}$ for all choices of scalars $\alpha_{ij}\in k,\ i,j=\overline{1,n}$. It's easy to check that this makes $\tilde\varphi$ a $k$-algebra homomorphism from $\mathcal M_{n}(k)$ into $\mathcal M_{m}(k)$.


Find all $n \in \mathbb{N}$ such that $3^{n}-n$ is divisible by $17$.

We have that $ 3^8\equiv(3^3)^2*3^2\equiv10^2*3^2\equiv( \minus{} 2)*3^2\equiv \minus{} 1\bmod{17}$so 3 is primitive root mod17.
Clearly 17 does not divide n.Let $ n\equiv{a}\bmod{17}$where $ a\in\left\{1,2,...16\right\}$.Because 3 is primitive root mod17 there is unique $ b\in\left\{1,2,...16\right\}$ such that $ 3^b\equiv{a}\bmod{17}$.
So $ 3^n\equiv{n}\equiv{a}\equiv3^b\bmod{17}\Rightarrow n\equiv{b}\bmod{16}$ We conclude that $ n\equiv x\bmod(16*17)$such that$ x\equiv a\bmod{17}$and$ x\equiv{b}\bmod{16}$.
This is the algorithm for finding the solutions mod272
We have $ 3\equiv3\mod{17},3^2\equiv9\bmod{17},3^3\equiv10\bmod{17},3^4\equiv13\bmod{17},3^5\equiv5\bmod{17},3^6\equiv15\bmod{17},3^7\equiv11\bmod{17},3^8\equiv16\bmod{17},3^9\equiv14\bmod{17},3^{10}\equiv8\bmod{17},3^{11}\equiv7\bmod{17},3^{12}\equiv4\bmod{17},3^{13}\equiv12\bmod{17},3^{14}\equiv2\bmod{17},3^{15}\equiv6\bmod{17},3^{16}\equiv1\bmod{17}$
If $ a \equal{} 1\Rightarrow b \equal{} 16\Rightarrow n\equiv256\bmod{272}$
If $ a \equal{} 2\Rightarrow b \equal{} 14\Rightarrow n\equiv205\bmod{272}$
If $ a \equal{} 3\Rightarrow b \equal{} 1\Rightarrow n\equiv241\bmod{272}$
If $ a \equal{} 4\Rightarrow b \equal{} 12\Rightarrow n\equiv140\bmod{272}$
If $ a \equal{} 5\Rightarrow b \equal{} 5\Rightarrow n\equiv5\bmod{272}$
If $ a \equal{} 6\Rightarrow b \equal{} 15\Rightarrow n\equiv159\bmod{272}$
If $ a \equal{} 7\Rightarrow b \equal{} 11\Rightarrow n\equiv75\bmod{272}$
If $ a \equal{} 8\Rightarrow b \equal{} 10\Rightarrow n\equiv42\bmod{272}$
If $ a \equal{} 9\Rightarrow b \equal{} 2\Rightarrow n\equiv162\bmod{272}$
If $ a \equal{} 10\Rightarrow b \equal{} 3\Rightarrow n\equiv163\bmod{272}$
If $ a \equal{} 11\Rightarrow b \equal{} 7\Rightarrow n\equiv215\bmod{272}$
If $ a \equal{} 12\Rightarrow b \equal{} 13\Rightarrow n\equiv29\bmod{272}$
If $ a \equal{} 13\Rightarrow b \equal{} 4\Rightarrow n\equiv132\bmod{272}$
If $ a \equal{} 14\Rightarrow b \equal{} 9\Rightarrow n\equiv201\bmod{272}$
If $ a \equal{} 15\Rightarrow b \equal{} 6\Rightarrow n\equiv134\bmod{272}$
If $ a \equal{} 16\Rightarrow b \equal{} 8\Rightarrow n\equiv152\bmod{272}$
(I correct my wrongs in calculations)

So what is your final answer? Not all those n's work, for example $ n \equal{} 3$ doesn't work?

all such $ n$ are in form $ n\equal{}16(17t\plus{}r\minus{}3^r)\plus{}r$ where $ r,t\in\mathbb Z$ which $ 0\leq r\leq 16$ and $ t\geq \frac{3^r\minus{}r}{17}$. A boring problem huh?  :maybe:


Prove that [tex]F_{n+1}^3-4F_n^3-F_{n-1}^3=3(-1)^nF_n[/tex], where [tex]\{F_n\}^{\infty}_{n=1}[/tex] is the well known Fibonacci sequence.


I'm having trouble getting the solutions template to work for me, so I'm going to send them each in one file.  However, I then have 5 + 5 = 10 files and Yahoo! only let's me attach 5 at a time, so can I just send 5 in each of two emails?

Also, I'm going to mail in my solutions as well, since the email is just a back up, and do I have to include the latex form of each solution as well as the printed pdf version or what?

[quote="vinny919"]I'm having trouble getting the solutions template to work for me, so I'm going to send them each in one file.  However, I then have 5 + 5 = 10 files and Yahoo! only let's me attach 5 at a time, so can I just send 5 in each of two emails?[/quote]
That's fine.  Just make sure that the subject lines of your emails say "1 of 2" and "2 of 2" or something like that, so we know to expect 2 messages.

[quote]Also, I'm going to mail in my solutions as well, since the email is just a back up, and do I have to include the latex form of each solution as well as the printed pdf version or what?[/quote]
If you're sending stuff through the regular mail, you just need to mail us your printed output (i.e. don't send us printed pages of LaTeX code, just send us the printouts of the PDF files).


Em muốn xin 1 số tài liệu hình học phẳng , vecto cũng đc ạ

Tài liệu giáo khoa chuyên toán lớp 10 phần Hinh học, trình bày có hệ thống, dễ hiểu, bài tập hay

Co mot it o day http://trungtuan.wordpress.com/2010/04/16/topic-42/


group theory
abstract algebra
inequalities
induction
Galois Theory
superior algebra
superior algebra unsolved

$ \text{1.}$Prove that a minimal normal subgroup $ N$ of a finite solvable group $ G$ is an elementary abelian $ p$-group for some prime $ p$.

$ \text{2.}$Let $ G$ be a group in which each subgroup is subnormal. If $ N$ is a normal subgroup of $ G$, $ Z(N)\neq1$, and $ G/N$ is cyclic then prove that $ Z(G)\neq1$

$ \text{3.}$If $ H$ is a normal subgroup of a group $ G$, and $ H$ and $ G/H^{\prime}$ are nilpotent groups of classes $ k$ and $ m$, respectively, then $ G$ is nilpotent of class $ \leq\left(2^{k}\minus{}1\right)m\minus{}2^{k\minus{}1}\plus{}1$.

1) Since $ G$ is solvable, we have $ N'\equal{}[N,N] < N$ is normal in $ G$ as a characteristic subgroup of $ N$. Therefore $ N$ must be abelian by the minimality of $ N$. Moreover, it must be a $ p$-group, since any $ p$-Sylowgroup of a finite abelian group is characteristic. It remains to show that $ N$ has exponent $ p$. This follows from the fact that the Frattini subgroup is a proper characteristic subgroup in a $ p$-group, ie. it must be equal to $ 1$, otherwise it would form a smaller minimal normal subgroup of $ G$.

2) Let $ G\equal{}<g>\cdot N$ and set $ U\equal{}<g>$. Since $ Z(N)$ is normal in $ G$ as a characteristic subgroup of $ G$, we can consider the subgroup $ H\equal{}U\cdot Z(N)$. The cyclic subgroup $ U$ is subnormal in $ G$ and hence, it is sub-normal in $ H$. In particular, $ U \subset N_H(U)\leq H$. We claim that $ N_H(U)\equal{}U\cdot N_{Z(N)}(U)$ holds with $ N_{Z(N)}(U)$ a normal subgroup of $ N_H(U)$. Clearly, $ N_H(U)\cap Z(N)\equal{}N_{Z(N)}(U)$ is a normal subgroup of $ N_H(U)$, therefore, it remains to show that $ N_H(U)\subseteq U\cdot N_{Z(N)}(U)$ holds. Let $ h\in N_H(U)\leq H$ be arbitrary. Then there exists some $ z\in Z(N)$ with $ h\in U\cdot z$. Since $ U\subseteq N_H(U)$, the element $ z$ must be contained in $ N_H(U)$ as well, i.e. $ z\in N_H(U)\cap Z(N)$. One consequence of the now verified claim is, that $ 1\not\equal{} N_{Z(N)}(U)\leq Z(N)$ holds. Note, that $ N_{Z(N)}(U)$ already commutes with $ N$. Since $ U$ and $ N_{Z(N)}(U)$ are both normal in $ N_H(U)$, we compute for their commutator $ C\equal{}[U,N_{Z(N)}(U)]\subseteq\, U\cap\, N_{Z(N)}(U)$. If $ C\equal{}1$ is valid, we have directly $ N_{Z(N)}(U)\subseteq Z(G)$. If $ C\not\equal{}1$, then $ C$ commutes with $ U$ and with $ N$, ie. $ C\subseteq Z(G)$.

3) It is trivial to show that $ \forall k\in \mathbb{N}$ the inequalities $ 2^{k\minus{}1}\geq k$ and $ 2^k > \frac{k(k\plus{}1)}{2}$ hold. E.g. for the second inequality we verify the claim easily for $ k\equal{}1,2,3$ and for $ k\plus{}1\geq 4$ we use induction, i.e. $ \frac{(k\plus{}1)(k\plus{}2)}{2}\equal{}\frac{k(k\plus{}1)}{2}\cdot \frac{k\plus{}2}{k}<2^k\cdot \frac{5}{3}<2^{k\plus{}1}$. Therefore, we have for all $ k,m\in\mathbb{N}$:

$ \left(2^k\minus{}1\minus{}\frac{k(k\plus{}1)}{2}\right)(m\minus{}1)\plus{}2^{k\minus{}1}\minus{}k\geq 0$ which is equivalent to $ (m\minus{}1)\frac{k(k\plus{}1)}{2}\plus{}k\leq m(2^k\minus{}1)\minus{}2^{k\minus{}1}\plus{}1$. By a famous result of P. Hall based on a repeated application of the Three Subgroup Lemma we know under the assumptions of the problem that $ G$ is nilpotent of class $ \leq (m\minus{}1)\frac{k(k\plus{}1)}{2}\plus{}k$. This completes the proof.


geometry
circumcircle
geometric transformation
reflection
Euler
cyclic quadrilateral
geometry solved

ABC is a triangle with orthocenter H. The median from A meets the circumcircle again at A1, and A2 is the reflection of A1 in the midpoint of BC. The points B2 and C2 are defined similarly. Show that H, A2, B2 and C2 lie on a circle.

Now I think G, the centroid, must also lie on the circle trough H, A2, B2, C2, but I don't know how to prove this.

In fact the center of the circle trough H, A2, B2, C2 should be the midpoint of [HG] (where G is the centroid). So it suffices to prove that A2G is perpendicular to A2H. I'm trying to prove this now.

Here?s a soln involving inversion (I just love inversion). Let?s make some notations: H1 is the foot of the altitude from A, H2 for B and H3 for C, M is the midpoint of BC.

Consider the inversion of pole A and power AH3*AB (=AH*AH1=AH2*AC, you can prove this easily). The circle H3H1MH2 (Euler?s circle) must be transformed in another circle which passes through B, C and H because these are the images of the pts H3, H2 and H1 respectively. The image of point M must also be on this circle (let?s call this image X). Because A, M and X are collinear we get that X is on the median AM (*). Because BHXC must be a cyclic quadrilateral angle BXC=angle BHC=pi-angle A=angle BA1C (**). From (*) and (**) we find that X is the reflection of A1 in the midpoint of BC, so X=A2=the image of pt M in the inversion we considered. This means that AH*AH1=AA2*AM, which in turn means that HH1MA2 is a cyclic quadrilateral, and because angle HH1M=90 deg we find angle HA2M=90 deg, so HA2 is perpendicular to the median from A=line A2G, which means that A2, B2, C2 are on a circle of diameter HG, exactly as you thought.

Does anyone have a solution without inversion ?
(Nice job Grobber, but I don't like inversion and I never see how to apply it the right way)
The problem simplified a bit. It suffices to prove the following.

Let ABC be a triangle with orthocenter H. Let F be the foot of the altitude which passes trough A. Let M be the midpoint of [BC], A(1) the intersection of AM with the circumcircle of ABC. Finally, let A(2) be the reflection of A(1) in M. Prove that A(2)H is perpendicular to AM.

One can try to prove that A(2)HFM is cyclic, or that triangles A(2)HA and AMF are similar. But up till now I didn't manage to do this ...

It is easy to see, that A_2, B_2, C_2 --- projections of H on the medians of ABC. So A_2, B_2,C_2,H, G lies on the circle, having GH as a diameter

[quote="Arne"] It suffices to prove the following.

Let ABC be a triangle with orthocenter H. Let F be the foot of the altitude which passes trough A. Let M be the midpoint of [BC], A(1) the intersection of AM with the circumcircle of ABC. Finally, let A(2) be the reflection of A(1) in M. Prove that A(2)H is perpendicular to AM.
[/quote]

Also simple solution
 prove that AHA2 ~ AMF (triangles) <=>
AM*AA2=AF*AH but AM*AA2=AM(AM-MA2)=AM(AM-MA1)=AM^2-P(M) where P(M) is the power of M. The rest is very simple computation.

and AF=2S/a and AH=2RcosA

fedor's is much more simple, but this is "valid" too


modular arithmetic
number theory unsolved
number theory

[quote="epsilon07"]Indeed problem seems to be easy but ı coludn't solve it :(

Think a number such that all of its digits are different from each other and also it is divisible by 11111   we call such numbers "interesting numbers". How many interesting number that consists of 10 different digits?[/quote]

The sum of the $ 10$ digits of such numbers is $ 0\plus{}1\plus{}2\plus{}3\plus{}4\plus{}5\plus{}6\plus{}7\plus{}8\plus{}9\equal{}45$ and so they are divisible by $ 9$, so by $ 99999$ since $ 11111\neq 0\pmod 3$

So, the required numbers may be written $ x(10^5\minus{}1)$ and $ x$ has exactly $ 5$ digits (else, $ x\equal{}100000$ or $ 100001$ and $ x(10^5\minus{}1)$ would be $ 9999900000$ or $ 9999999999$ and could not have all its digits different).

It's then immediate to see that $ x(10^5\minus{}1)\equal{}\overline{abcde(9\minus{}a)(9\minus{}b)(9\minus{}c)(9\minus{}d)(9\minus{}e)}$ for some $ a,b,c,d,e$ and so :

$ a$ may take $ 9$ different values (I suppose that when the problem says "$ 10$ digits numbers", the leftmost one is non-zero)
$ b$ may then take any of the $ 8$ remaining values (we exclude $ a$ and $ 9\minus{}a$)
$ c$ may then take any of the $ 6$ remaining values
$ d$ may then take any of the $ 4$ remaining values
$ e$ may then take any of the $ 2$ remaining values

And the result is $ 9.8.6.4.2\equal{}3456$ such numbers.

very well thanks a lot nice solution....

Let $n=\overline{d_1d_2...d_{10}}$ be a 10-digit interesting number. Note that $\{d_i\}_{i=1}^{10}$ are $\{i\}_{i=0}^9$ in some order, and obviously $d_1\not=0$. This means $\sum_{i=1}^{10}d_i\equiv 45\equiv 0\pmod 9$ so $9\mid n$. Since and $\gcd(9,11111)=1$, and we also know that any interesting number must be divisible by $11111$, it follows that $9(11111)=99999\mid n$. Now let $x=\overline{d_!d_2d_3d_4d_5}$ and let $y=\overline{d_6d_7d_8d_9d_{10}}$, so that $n=10^5x+y\equiv x+y\equiv 0\pmod {99999}$. But $x+y<99999+99999=2(99999)$ so in order for the congruence to be true we must have $x+y=(0,99999)$. The first one is clearly false so $x+y=99999$. This means that $d_i+d_{i+5}=9$ for $i=1,3,5,7,9$. There are $5!=120$ ways to distribute $(0,9),(1,8),...,(4,5)$ among $(d_1,d_6),(d_3,d_8),(d_4,d_9),(d_5,d_{10})$, and for each pair the numbers are interchangeable (e.g. $(d_2,d_7)$ can be $(0,9)$ [i]as well as[/i] $(9,0)$) in $2^5=32$ ways for a total of $32(120)=3840$ ways. However, we have over-counted a $\tfrac{1}{10}$th of these numbers which have $a=0$. This is clearly not allowed, so there are actually $3840-\tfrac{1}{10}(3840)=\tfrac{9}{10}(3840)=9(384)=\boxed{3456}$ interesting numbers.

[quote="Mathematicalx"]I cant remember the year but thats from Turkish first round.[/quote]
Well actually the original source is an old Russia MO.


Prove that $ 21 | 4^{n\plus{}1} \plus{} 5^{2n\minus{}1}$ for every natural number $ n$.

I think I have it with mod 3 and mod 7, but I don't understand how you'd use induction to prove it...o_O


[u][b]Problem 1:[/b][/u] Find all reals $ (x,y)$ satisfying
\begin{align} x^2 + 4y & = 21 \\
y^2 + 4x & = 21 \end{align}

[hide="Answer"]$ (3,3),( - 7, - 7),( - 1,5),(5 - 1)$[/hide]

[hide="Hint"]Try to build squares.[/hide]

[hide="Solution"]
(1)-(2):$ x^2 - 4x - (y^2 - 4y) = 0\Rightarrow x^2 - 4x + 4 - (y^2 - 4y - 4) = 0\Rightarrow \\
(x + 2)^2 - (y + 2)^2 = 0\Rightarrow (x - y)(x + y - 4) = 0$
$ x = y$ yields $ (3,3),( - 7, - 7)$, $ x + y = 4$ yields $ ( - 1,5),(5, - 1)$[/hide]

[u][b]Problem 2:[/b][/u]The excircle $ c$ of an isosceles triangle $ \Delta ABC$ $ \left(|\overline{AC}| = |\overline{BC}|\right)$ touches $ \overline{BC}$ internally. Prove that its radius is the same as the height from $ C$ to $ \overline{AB}$.

[hide="Hint"]Use $ A = \frac {(b + c - a)r_a}2.$[/hide]

[hide="Answer"]$ b = a\Rightarrow \frac {ch_c}2 = A = \frac {b + c - a)r_a}2 = \frac {cr_a}2\;\;(a = b).$[/hide]

[u][b]Problem 3:[/b][/u] Is there a magic square that starts with the numbers in the attachment?

[hide="Answer"]No.[/hide]

[hide="Hint"]Prove that there has to be a 13 in the top left corner, which is impossible.[/hide]

[hide="Solution"]We can find that on top of the 13 is a 1, on top of the 15 is a 5 and in the top right corner there's a 16. Now look at the different numbers we can put into the bottom left corner (2 and 3). No matter what we put there, we'll have to write a 13 in the top left corner.[/hide]

[u][b]Problem 4:[/b][/u]We call a positive integer $ n$ [i]stout[/i] if there are exactly 2 pairs of positive integers $ (k,l)$ with $ \frac 1n = \frac 1k + \frac 1l$ and $ k\le l$. Prove that $ n$ is stout iff $ n$ is a prime.

[hide="Hint"]$ \frac 1n = \frac 1{2n} + \frac 1{2n} = \frac 1{n + 1} + \frac 1{n(n + 1)}$ and $ \frac 1n = \frac 1k + \frac 1l \Rightarrow \frac 1{mn} = \frac 1{mk} + \frac 1{ml} = \frac 1{k'} + \frac 1{l'}.$[/hide]

[hide="Solution"]
First, we observe $ n < k\le2n\le l$, and a direct condequence is that 1 is not stout as $ k$ has to be 2. Now consider a non-prime number $ n$. Per definitionem $ n$ has at least one divisor $ 1 < x < n$. Then $ 1 < \frac nx < n$ is a natural number, too. There are at least 2 ways to write $ x$ as $ \frac 1k + \frac 1l$ (look it up in the hint). Then for each such pair we have $ \frac 1n = \frac 1{xy} = \frac 1{ky} + \frac 1{ly}$, which are at least two compositions. But for each $ (k',l')$ We found this way $ \gcd(n,k') = \gcd(xy,ky) \neq 1$ and $ \gcd(n,l')\neq 1$. However, $ \gcd(n,n + 1) = 1$ wo we have to have another pair $ (n + 1,n(n + 1))$ which makes more than two pairs in total so $ n$ isn't stout.

Now let $ p$ be a prime number and $ \frac 1p = \frac 1k + \frac 1l \Leftrightarrow p = \frac {kl}{k + l}\Leftrightarrow p|kl\Leftrightarrow p|k \or p|l.$
[u]Case 1: [/u]$ p|k$
$ p < k\le 2p\Rightarrow k = 2p \Rightarrow (k,l) = (2p,2p)$
[u]Case 2: [/u]$ p|l$
Write $ l = qp$; then $ \frac 1p = \frac 1k + \frac 1{qp} \Rightarrow k = \frac {pq}{q - 1}\in\mathbb{N}.$ Now $ gcd(q,q - 1) = 1$ so we have to have $ (q - 1)|p\Rightarrow q = 2 \or q = p + 1.$ $ q = 2$ yields $ (2p,2p)$ and $ q = p + 1$ yields $ (p + 1,p(p + 1))$ which are all solutions. So we conclude that if $ n$ is prime it is stout and the problem is solved.
[/hide]

No. 4
[hide]
$ \frac {1}{n} \equal{} \frac {1}{k} \plus{} \frac {1}{l}$
$ \Rightarrow n \equal{} \frac {kl}{k \plus{} l}$
$ \Rightarrow kl \minus{} n(k \plus{} l) \equal{} 0$
$ \Rightarrow (k \minus{} n)(l \minus{} n) \equal{} n^2$
Because $ k, l, n$ are all positive integers, there is,
$ k \minus{} n \equal{} p$
$ l \minus{} n \equal{} q$
where $ p,q$ are integers and $ pq \equal{} n^2$; because $ k \leq l, p\leq q$. Then,
$ (k, l) \equal{} (n \plus{} p, n \plus{} q)$
Because $ k > 0, l > 0 \Rightarrow n > \minus{} p, n > \minus{} q.$
If $ p < 0 \Rightarrow q < 0 \Rightarrow n > |p|, n > |q| \Rightarrow n^2 > |pq| \equal{} pq$. Contradiction. Therefore, $ p > 0, q > 0.$

[b]Case 1[/b]: $ n \equal{} 1$. Then the only positive integer factoring of $ n^2 \equal{} 1$ is $ (p, q) \equal{} (1, 1)$. Only one solution.

[b]Case 2[/b]: $ n$ is a prime. Then all the $ (p, q)$ are $ (1, n^2)$ and $ (n, n)$. Exactly two solutions.

[b]Case 3[/b]: $ n$ is a composite. Let $ n \equal{} a\cdot m$, where $ a$ is a prime and $ m$ is not 1. Then besides the factorings mentioned in Case 2, there's at least one more factoring $ (a, a\cdot m^2)$. More than two solutions.

Hence, iff $ n$ is a prime, the statement holds.[/hide]

[hide="1"]We add the equations and add two fours:

$ x^2\plus{}4x\plus{}4\plus{}y^2\plus{}4y\plus{}4\equal{}50$

$ (x\plus{}2)^2\plus{}(y\plus{}2)^2\equal{}50$

We can easily graph that circle.[/hide]

@bertram: The problem said reals, not integers.

[quote="1=2"][hide="1"]We add the equations and add two fours:

$ x^2 \plus{} 4x \plus{} 4 \plus{} y^2 \plus{} 4y \plus{} 4 \equal{} 50$

$ (x \plus{} 2)^2 \plus{} (y \plus{} 2)^2 \equal{} 50$

We can easily graph that circle.[/hide]

@bertram: The problem said reals, not integers.[/quote]Check your work before posting. Not all real (in fact almost none) points in that circle [u]satisfy the original system[/u]

[hide="1"] Adding the equations gives $ (x \plus{} 2)^2 \plus{} (y \plus{} 2)^2 \equal{} 50$.

Subtracting, we get

$ x^2 \minus{} y^2 \plus{} 4y \minus{} 4x \equal{} 0 \Leftrightarrow$
$ (x \minus{} y)(x \plus{} y \minus{} 4) \equal{} 0$

When $ x \equal{} y$ we have $ (x \plus{} 2)^2 \equal{} 25$, so $ (x, y) \equal{} \boxed{ (3, 3), (\minus{}7, \minus{}7) }$

When $ 4 \minus{} x \equal{} y$ we have

$ x^2 \plus{} 4x \plus{} 4 \plus{} 36 \minus{} 12x \plus{} x^2 \equal{} 2x^2 \minus{} 8x \plus{} 40 \equal{} 50 \Leftrightarrow$
$ x^2 \minus{} 4x \minus{} 5 \equal{} 0 \Leftrightarrow$
$ (x \minus{} 5)(x \plus{} 1) \equal{} 0$

So $ (x, y) \equal{} \boxed{ (5, \minus{}1), (\minus{}1, 5) }$.

Generally, two parabolas intersect in four points.   [/hide]

[quote="t0rajir0u"][hide="1"] Adding the equations gives $ (x \plus{} 2)^2 \plus{} (y \plus{} 2)^2 \equal{} 50$.

Subtracting, we get

$ x^2 \minus{} y^2 \plus{} 4y \minus{} 4x \equal{} 0 \Leftrightarrow$
$ (x \minus{} y)(x \plus{} y \minus{} 4) \equal{} 0$

When $ x \equal{} y$ we have $ (x \plus{} 2)^2 \equal{} 25$, so $ (x, y) \equal{} \boxed{ (3, 3), ( \minus{} 7, \minus{} 7) }$

When $ 4 \minus{} x \equal{} y$ we have

$ x^2 \plus{} 4x \plus{} 4 \plus{} 36 \minus{} 12x \plus{} x^2 \equal{} 2x^2 \minus{} 8x \plus{} 40 \equal{} 50 \Leftrightarrow$
$ x^2 \minus{} 4x \minus{} 5 \equal{} 0 \Leftrightarrow$
$ (x \minus{} 5)(x \plus{} 1) \equal{} 0$

So $ (x, y) \equal{} \boxed{ (5, \minus{} 1), ( \minus{} 1, 5) }$.

Generally, two parabolas intersect in four points.   [/hide][/quote]
If this problem was on an olympiad, could we find those four solutions and assert that since two distinct parabolas have at most four intersections points, we're done?

[quote="t0rajir0u"]The solution I gave is complete without that last statement.  Alternately, you could cite Bezout's Theorem, which in real affine space states that the [b]maximum[/b] number of intersections of a curve of degree $ n$ and a curve of degree $ m$ is $ mn$.[/quote]

You'd have to prove this theorem though, and that's definitely not worth it. :)

You could prove it for this problem by substituting $ y \equal{} \frac{21\minus{}x^2}4$ and obtain an equation of degree 4, of which you can guess the roots -7, -1, 3, 5 and then compute the corresponding y's.

[quote="t0rajir0u"][quote="bertram"][quote="t0rajir0u"]The solution I gave is complete without that last statement.  Alternately, you could cite Bezout's Theorem, which in real affine space states that the [b]maximum[/b] number of intersections of a curve of degree $ n$ and a curve of degree $ m$ is $ mn$.[/quote]

You'd have to prove this theorem though[/quote]

Bezout's Theorem is an extremely well-known result.  As long as it's cited, it doesn't need to be proven, and the proof of the general theorem is much more difficult.[/quote]

Actually, there's a disclaimer on top of the page with the problems that sais you're only allowed to use theorems if you know them from class. Another example where the problem would have been made trivial if you applied a well known theorem is finding all natural numbers x and y with $ (x\minus{}1)^{2y}\plus{}x^{2y} \equal{} (x\plus{}1)^{2y}$ which is extremely easy once you use fermat's last theorem... it's from the german imo qualification exams 1998, so you's guess you can't just prove it that easily ;)


abstract algebra
group theory
superior algebra
superior algebra unsolved

suppose$ D$ be a division ring.$ D^* \equal{}D\minus{}${0 } as the definition is a group with multiplication action and necesserily is abelian.show if $ charD\equal{}p>0$ then every finite and abelian subgroup of $ D^*$ is cyclic.

Er, do you mean *not* necessarily abelian?

EDIT: nevermind, maybe it's true, but I don't see it.

If $ D$ is a division ring and $ G\le D^\times$ finite, then $ G$ must be cyclic if either $ G$ abelian [b]or[/b] $ \text{char} D\equal{}p>0$.
Look here [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=201470[/url].

yes.it is not necessrilly abelian.the correct question was wrote by orilin.