geometry
3D geometry
algebra solved
algebra

The sequence {a_n,k} = 1, 2,..., 2^n; n = 0,1,2,... is defined by

(1) a_0,1 = 2,
(2) a_n,k = 2 * (a_(n-1),k)^3 and
(3) a_n,(k+2^(n-1)) = 1/2 * (a_(n-1),k)^3.

Prove that all a_n,k are distinct.

[i]German IMO selection test 1979 problem 6.[/i]

I didn't check if my solution is wrong, but I think that the idea is the following:
  Suppose that there exists m,n,k,p such that a_nk=a_mp. Let us take m and n such that m+n is minimum with the property that there exists k<=2^n and p<=2^m with a_nk=a_mp. We have three cases:
  Case 1
  If k<=2^n-1 and p<=2^m-1, then it follows that 2a^3_n-1,k=2a^3_m-1,p and so there exits k<=2^n-1 and p<=2^m-1 with a_n-1,k=a_m-1,p, contradicting the minimality of m+n.
  Case 2
   If k>2^n-1 and p>2^n-1, then write k=k_1+2^n-1 and p=p_1+2^m-1. From the recursive relation it follows that a_n-1,k_1=a_m-1,p_1 and k_1<=2^n-1, p_1<=2^m-1, contradiction.
  Case 3
   If for example k>2^n-1 and p<=2^m-1 then it follows that 4 is a perfect cube, false.


Examine the following list of events that occur during the propagation of a nerve impulse:

I - Neurotransmitters reach the dendrites
II - Neurotransmitters are released by the end of the axon.
III - The impulse spreads the axon.
IV - The impulse spreads by dendrites.
V - The impulse reaches the cell body.

Alternative that presents the chronology of events correct ?

a) V-III-I-IV-II
b) I-IV-V-III-II
c) I-IV-III-II-V
d) II-I-IV-III-V
e) II-III-I-IV-V

Thanks.

the answer shud b (e)

but first check out if ur (III) is gramatically correct :)


[quote="Michael Niland"]A point is taken on each side of a parallelogram and the points on adjacent sides are joined by line segments so as to form a quadrilateral.
If the area of this quadrilateral is half the area of the parallelogram prove that at least one diagonal of the quadrilateral is parallel to a side of the parallelogram[/quote]

[b]HINT[/b]

 If $LN$ is not parallel to $AD,BC$, then we draw $MR \parallel AB$ , so $( RLMN) = \frac{1}{2}(ABCD) =(KLMN)$.
 This means that $(KLN) = (KRN)$ and hence $KR \parallel LN$ , contradiction!. So points $K,R$ coincide , that is $MK \parallel AB$


[quote="Shinraehee"]Let $ a,b,c > 0$. Prove that:
$ \sqrt {(a^2 b \plus{} b^2 c \plus{} c^2 a)(b^2 a \plus{} c^2 b \plus{} a^2 c)} \ge abc \plus{} \sqrt [3]{(a^3 \plus{} abc) \plus{} (b^3 \plus{} abc) \plus{} (c^3 \plus{} abc)}$[/quote]
I think it's should be \[ \sqrt {(a^2 b \plus{} b^2 c \plus{} c^2 a)(b^2 a \plus{} c^2 b \plus{} a^2 c)} \ge abc \plus{} \sqrt [3]{(a^3 \plus{} abc)(b^3 \plus{} abc)(c^3 \plus{} abc)}\]
See here: http://www.mathlinks.ro/viewtopic.php?t=166168


The molecular weight of an unknown substance was determined to be approximately 94 g/mol. Knowing that each molecule of the substance contains a single oxygen atom, identify the substance.

phenol in the monomer state among others

And what was the method you used to arrive at that conclusion?

Actually phenol is the only possible answer, but the purpose was not to guess.

Carcul,can you post more of those ABCD type questions in organic and inorganic?? (please)

subtract 16 from 94 to get 78. now consider all compounds with the weight 78 (C6H6 being one of them) divide by 12 to get 6 which corr to the no of c atmos and and the rest are hydrogen and you get the the formula c6h6. Now other compounds could include nitrogen, but the no of nitrogen atoms cannot be odd as that would lead to a fractional du and hence impossible compound(for eg. C5NOH4) presence of two or more nitrogen would lead to a compound with more hydrogen atoms than permissible and also a fractional DU(for eg C4N2OH2 or C3N2OH14). Similarly you can test the presence of other elements. This Is done of course on the assumption that only normal isotopes are present in the compound(for eg, you can have a compound with formula C5NOH2D or C3N2OD7 )so only molecules with formula C6H6O have mass 94

Well, this is in fact a basic problem of spectroscopy. For a given molar mass M of an organic compound, we start by writting M in the form

$ \frac{M}{13} \equal{} n \plus{} \frac{m}{13}$,

and we then consider our "base formula" as $ C_nH_{n \plus{} m}$. In our case we get n = 7 and m = 3, and so our base formula will be $ C_7H_{10}$. As we have an oxygen, then for adding it we must subtract for the base formula a combination of C and H with the same mass: $ CH_4$. After doing this, we get the formula of our compound $ C_6H_6O$. The most stable compound with this formula is phenol.

The reasoning I gave above is known as the "Rule of Thirteen". Here's one more example:

-Find the molecular formulas for possible organic compounds with molecular masses of 136, in the following cases:

(a) A compound with two oxygen atoms.
(b) A compound with two nitrogen atoms.
(c) A compound with two nitrogen atoms and one oxygen atom.
(d) A compound with five carbon atoms and four oxygen atoms.

(For those interested, there are still 3 more interesting problems like this.)

Yes, that's right. Some possible structures?


ratio
geometry
perimeter
LaTeX
Pythagorean Theorem

You guys will probably need a diagram for this...

A square ABCD has line segmends drawn from vertex B to the midpoints N and M of sides AD and DC respectively. Find the ratio of the perimeter of quadrilateral BMDN to the perimeter of square ABCD.

Want a hint? :)

[hide]Use the Pythagorean Theorem to find some side lengths of quadrilateral BMDN.[/hide]

Funny how you can draw a diagram and realize it probably didn't help you very much at all...
[hide="solution"]let the side length of the square be 2x. the perimeter is 8x
ND and MD are x, and by pythagorean theorem BN and BM are xsqrt5.
(x+x+xsqrt5+xsqrt5)/8x simplifies to (1+sqrt5)/4
[/hide]

[hide]Let the side length be 2. $BN=BM=\sqrt{5}$.
Perimeter of BMDN is $2+2\sqrt{5}$.
Perimeter of ABCD is 8.
The ratio is $\frac{2+2\sqrt{5}}{8}=\frac{1+\sqrt{5}}{4}$.[/hide]

[quote]Funny how you can draw a diagram and realize it probably didn't help you very much at all... [/quote]

Well, when I drew one, it wuz like my life depended on it...lol :rotfl:

[hide]It works out to be$(x(sq5)+x)/4x=(1+sq(5))/4$[/hide]

diagram helped me a lot >_>
[hide]
$\frac{1 + \sqrt{5}}{4}$
[/hide]

[quote="Knuth's_check"]Let the side be 4.  [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/8edac681582cee5938f72fe0bcb232a4.gif[/img], therefore[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/3e1015a8640b40a4abccacf8fc2bc7ab.gif[/img]. Because BN=BM, perimeter ratio would be[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/f50d415280176822a4014a3a2843e591.gif[/img][/quote]
[hide]$\frac{1+\sqrt{5}}{4}$[/hide]

[quote="Knuth's_check"][hide]Let the side be 4.  [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/8edac681582cee5938f72fe0bcb232a4.gif[/img], therefore[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/3e1015a8640b40a4abccacf8fc2bc7ab.gif[/img]. Because BN=BM, perimeter ratio would be[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/f50d415280176822a4014a3a2843e591.gif[/img] [/hide][/quote]

1. it can be simplified  ;)
2. Please hide your answers and solutions so others will have a chance to do the problems.
3. I think you are using the TeXeR to type up things in LaTeX. If you want to use LaTeX on the forums, just put your code in between two dollar signs.


algebra
IMO Shortlist
functional equation

[quote="vietnamesegauss89"]
$\Rightarrow 4f(x)+3+(f(x)-2x+1)(f(x)+2x+1)=f(x^2)$
[/quote]
How you get that ?
I get  $4f(x)+3+(f(x)-2x+1)(f(x)+2x+1)=f(-x^2)$

Anyone help me please!

can anyone explain how he could get :

Given $y=-1-x$ we have , $f(x)(f(x)+2x+1)=f(x^2+x-1)$

thanks

Hi Here you have found as follows
[quote="vietnamesegauss89"]Given$y=-1-x$we have:
$f(x)(f(x)+2x+1)=f(x^2+x-1)$
Given$x=1$we have:
$f(1)=f(1)(f(1)+3)$
$\Rightarrow f(1)=0$or$f(1)=-2$[/quote]
$\Rightarrow f(1)=0$or$f(1)=-2$. you used "or"  but it could be the same or I am wrong. explain it how to discover that they are different
Abdurashid

They are allowed to be the same; the or is not exclusive in that scenario.

Here's my solution:

Let $ P(x,y)$ be the assertion that $ f(x \plus{} y) \plus{} f(x)f(y) \equal{} f(xy) \plus{} 2xy \plus{} 1$.

$ P(0,0)$ yields $ f(0) \equal{} \pm 1$. If $ f(0) \equal{} 1$, then $ P(x,0)$ yields $ f(x) \equal{} 1$ for all $ x$, which $ P(3,3)$ shows is impossible, so $ f(0) \equal{} \minus{} 1$.

$ P(x, \minus{} x)$ now gives $ f(x)f( \minus{} x) \equal{} f( \minus{} x^2) \minus{} 2x^2 \plus{} 2$. Setting $ x \equal{} 1$ yields $ f(1)f( \minus{} 1) \equal{} f( \minus{} 1)$, so $ f(1) \equal{} 1$ or $ f( \minus{} 1) \equal{} 0$.

If $ f(1) \equal{} 1$, then $ P(x,1)$ yields $ f(x \plus{} 1) \plus{} f(x) \equal{} f(x) \plus{} 2x \plus{} 1$, so $ f(x \plus{} 1) \equal{} 2x \plus{} 1$, so $ f(x) \equal{} 2x \minus{} 1$ for all real $ x$. It can easily be seen that $ f(x) \equal{} 2x \minus{} 1$ indeed satisfies this functional equation.

Suppose now that $ f(1) \neq 1$; then $ f( \minus{} 1) \equal{} 0$. $ P(1, \minus{} 2)$ yields $ f( \minus{} 1) \plus{} f(1)f( \minus{} 2) \equal{} f( \minus{} 2) \minus{} 3$, that is, $ f(1)f( \minus{} 2) \equal{} f( \minus{} 2) \minus{} 3$. $ P( \minus{} 1,2)$ gives $ f(1) \plus{} f( \minus{} 1)f(2) \equal{} f( \minus{} 2) \minus{} 3$, that is, $ f(1) \equal{} f( \minus{} 2) \minus{} 3$. Therefore, $ f(1) \equal{} f(1)f( \minus{} 2)$, so $ f(1) \equal{} 0$ or $ f( \minus{} 2) \equal{} 1$.

$ f(1) \equal{} 0$ will give $ f(x) \equal{} x^2 \minus{} 1$. $ P(x, \minus{} 1)$ and $ P( \minus{} x,1)$ yield $ f(x \minus{} 1) \equal{} f(1 \minus{} x) \equal{} f( \minus{} x) \minus{} 2x \plus{} 1$, so $ f$ is even. $ P(x, \minus{} x)$, combined with the fact that $ f$ is even, yields $ f(x)^2 \equal{} f(x^2) \minus{} 2x^2 \plus{} 2$, while $ P(x,x)$ gives $ f(2x) \plus{} f(x)^2 \equal{} f(x^2) \plus{} 2x^2 \plus{} 1$. Since $ f(x)^2 \equal{} f(x^2) \minus{} 2x^2 \plus{} 2$, we see that $ f(2x) \equal{} 4x^2 \minus{} 1$, so $ f(x) \equal{} x^2 \minus{} 1$ for all $ x$; it can easily be verified that $ f(x) \equal{} x^2 \minus{} 1$ satisfies the functional equation.

If $ f( \minus{} 2) \equal{} 1$, then $ f(1) \equal{} f( \minus{} 2) \minus{} 3 \equal{} \minus{} 2$. $ P(1, \minus{} x)$ yields $ f(1 \minus{} x) \equal{} 3f( \minus{} x) \minus{} 2x \plus{} 1$, while $ P( \minus{} 1,x)$ yields $ f(x \minus{} 1) \equal{} f( \minus{} x) \minus{} 2x \plus{} 1$. Subtracting these two equations gives $ 2f( \minus{} x) \equal{} f(1 \minus{} x) \minus{} f(x \minus{} 1)$. Substituting $ x \equal{} 1 \minus{} y$ yields $ 2f( \minus{} 1 \plus{} y) \equal{} f(y) \minus{} f( \minus{} y)$. Replacing $ y$ with $ \minus{} y$ yields $ 2f( \minus{} 1 \minus{} y) \equal{} f( \minus{} y) \minus{} f(y) \equal{} \minus{} 2f( \minus{} 1 \plus{} y)$, so $ f( \minus{} 1 \plus{} y) \equal{} \minus{} f( \minus{} 1 \minus{} y)$ for all real $ y$.

$ P(x, \minus{} 1)$ yields $ f(x \minus{} 1) \equal{} f( \minus{} x) \minus{} 2x \plus{} 1$, and $ P( \minus{} x, \minus{} 1)$ yields $ f( \minus{} x \minus{} 1) \equal{} f(x) \plus{} 2x \plus{} 1$. Since $ f( \minus{} x \minus{} 1) \equal{} \minus{} f(x \minus{} 1)$, we have that $ f(x \minus{} 1) \equal{} \minus{} f(x) \minus{} 2x \minus{} 1 \equal{} f( \minus{} x) \minus{} 2x \plus{} 1$, so $ f(x) \plus{} f( \minus{} x) \equal{} \minus{} 2$ for all real $ x$.

$ P(0,0)$ yields $ f(x)f( \minus{} x) \equal{} f( \minus{} x^2) \minus{} 2x^2 \plus{} 2$, so $ f(x)( \minus{} 2 \minus{} f(x)) \equal{} \minus{} 2 \minus{} f(x^2) \minus{} 2x^2 \plus{} 2$. Thus, $ 2f(x) \plus{} f(x)^2 \equal{} f(x^2) \plus{} 2x^2$. On the other hand, $ P(x,x)$ yields $ f(2x) \plus{} f(x)^2 \equal{} f(x^2) \plus{} 2x^2 \plus{} 1$. Subtracting $ 2f(x) \plus{} f(x)^2 \equal{} f(x)^2 \plus{} 2x^2$ gives $ f(2x) \minus{} 2f(x) \equal{} 1$.

$ f(2) \equal{} \minus{} 2 \minus{} f( \minus{} 2) \equal{} \minus{} 3$, so $ P(x,2)$ yields $ f(x \plus{} 2) \minus{} 3f(x) \equal{} f(2x) \plus{} 4x \plus{} 1$. But $ f(2x) \equal{} 2f(x) \plus{} 1$, so $ f(x \plus{} 2) \equal{} 5f(x) \plus{} 4x \plus{} 2$. On the other hand, $ P(x,1)$ gives $ f(x \plus{} 1) \equal{} 3f(x) \plus{} 2x \plus{} 1$; substituting $ x \equal{} y \plus{} 1$ here yields $ f(y \plus{} 2) \equal{} 3f(y \plus{} 1) \plus{} 2y \plus{} 3$. But $ f(y \plus{} 1) \equal{} 3f(y) \plus{} 2y \plus{} 1$, so $ f(y \plus{} 2) \equal{} 9f(y) \plus{} 8y \plus{} 6$. However, $ f(y \plus{} 2) \equal{} 5f(y) \plus{} 4y \plus{} 2$ as well, so $ 4f(y) \equal{} \minus{} 4y \minus{} 4$, yielding $ f(y) \equal{} \minus{} y \minus{} 1$ for all real $ y$. It can easily be verified that $ f(x) \equal{} \minus{} x \minus{} 1$ satisfies the functional equation.

Hence, our three solutions to this functional equation are $ \boxed{f(x) \equal{} 2x \minus{} 1, x^2 \minus{} 1, \minus{} x \minus{} 1}$.

This is my solution from WOOT some time ago...

Let $P(x,y)\implies f(x+y)+f(x)f(y)=f(xy)+2xy+1$. First, note that $P(1,-1)\implies f(-1)[f(1)-1]=0$.

[b]Case 1:[/b] $f(1)=1$. Then
\[P(x,1)\implies f(x+1)=2x+1\implies f(x)=2x-1\]for all $x$.

[b]Case 2:[/b] $f(1)\ne1\implies f(-1)=0$. Note that $P(x,0)\implies f(0)=-1$ (otherwise, $f(x)$ is constant, which is clearly impossible since $2xy$ is surjective). Now let $c=1-f(1)$ (by assumption, $c\ne0$). Then
\[P(x,1)\implies f(x+1)=cf(x)+2x+1.\]The second order difference gives us
\[f(x+3)-(c+2)f(x+2)+(2c+1)f(x+1)-cf(x)=0\]for all $x$.

[b]Subcase 2.1:[/b] If $c=1$ (i.e. $f(1)=0$), so $f(x+1)=f(x)+2x+1$ for all $x$ and we find by simple induction that $f(n)=n^2-1$ and $f(x+n)=f(x)+(x+n)^2-x^2$ for all integers $n$, whence
\[P(x,\pm n)\implies f(-xn)=f(xn)+1=n^2[f(x)+1].\]Thus $f$ is even, and
\[P(x,x)\implies f(x^2)=[f(x)]^2+4f(x)-2x^2+2\]while
\[P(x,-x)\implies f(x^2)=f(-x^2)=[f(x)]^2+2x^2-2.\]Equating, we arrive at
\[f(x)=x^2-1,\]which indeed satisfies the original equation.

[b]Subcase 2.2:[/b] If $c\ne1$ (i.e. $f(1)\ne0$), then the characteristic polynomial of the sequence $f(n)$ has roots $1,1,c$ and
\[f(n)=\alpha+\beta n+\gamma c^n\]for all integers $n$ (for some real constants $\alpha,\beta,\gamma$). Considering $f(-1),f(0),f(1)$, we find that
\[0=\alpha-\beta+\frac{\gamma}{c},\quad -1=\alpha+\gamma,\quad 1-c=\alpha+\beta+\gamma c.\]Solving this system,
\[f(n)=-\frac{c+1}{(c-1)^2}+\frac{2n}{1-c}-\frac{c(c-3)}{(c-1)^2}c^n.\]Now,
\[P(2,2)\implies f(2)=\pm3.\]If $f(2)=3$, then
\[3=f(2)=cf(1)+2(1)+1=cf(1)+3\implies cf(1)=0.\]But we have $c\ne0$ and $f(1)\ne0$. So $f(2)=-3$, and
\[-3=f(2)=cf(1)+2(1)+1\implies [f(1)-3][f(1)+2]=0.\]
[b]Subcase 2.2.1:[/b] $f(1)=3\implies c=-2$. Then
\[f(n)=\frac{1}{9}+\frac{2n}{3}-\frac{10}{9}(-2)^n\]for all integers $n$, so $f(2)=-3$, $f(-2)=-3/2$, and $f(-4)=-21/8$. This contradicts $P(2,-2)\implies f(2)f(-2)=f(-2^2)$.

[b]Subcase 2.2.2:[/b] $f(1)=-2\implies c=3$. Then
\[f(n)=-1-n\]for all integers $n$. By induction, we find $f(x+n)=3^nf(x)+(3^n-1)x+(3^n-n-1)$. Thus
\[P(x,n)\implies f(xn)+xn+1=(3^n-n-1)[f(x)+x+1].\]So
\begin{align*}
(3^n-n-1)^2[f(x)+x+1]&=(3^n-n-1)[f(xn)+xn+1]\\
&=f(xn^2)+xn^2+1=(3^{n^2}-n^2-1)[f(x)+x+1]
\end{align*}for all real $x$ and integers $n$. Take $n=-1$. We find
\[f(x)=-x-1.\]
Finally, our solutions are $f(x)=2x-1\forall x$, $f(x)=x^2-1\forall x$, and $f(x)=-x-1\forall x$ (all three work).

[quote="andre.l"]Find all functions $ f: \mathbb{R}\to\mathbb{R}$ such that $ f\left(x+y\right)+f\left(x\right)f\left(y\right)=f\left(xy\right)+2xy+1$ for all real numbers $ x$ and $ y$.[/quote]
Let $P(x,y)$ be the assertion $f(x+y)+f(x)f(y)=f(xy)+2xy+1$
Let $f(0)=a$
Let $f(1)=b$
Let $f(-1)=c$

(e1) : $P(x,-1)$ $\implies$ $f(x-1)+cf(x)=f(-x)-2x+1$
(e2) : $P(x-1,1)$ $\implies$ $f(x)+(b-1)f(x-1)=2x-1$
(1-b)e1 + e2 $\implies$ $(1+c-bc)f(x)+(b-1)f(-x)=-2bx-b$
changing $x\to -x$, we get $(1+c-bc)f(-x)+(b-1)f(x)=2bx-b$

And so the system :
$(1+c-bc)f(x)+(b-1)f(-x)=-2bx-b$
$(b-1)f(x)+(1+c-bc)f(-x)=2bx-b$

If the determinant of the system is non zero, this gives $f(x)=ux+v$ for some real $u,v$ and plugging this in original equation,
we get two solutions $\boxed{f(x)=-x-1}$ and $\boxed{f(x)=2x-1}$

If the determinant of the system is zero, this means :
either $1+c-bc=b-1$ and so $-2bx-b=2bx-b$ and so $b=0$
either $1+c-bc=1-b$ and so $-2bx-b=-(2bx-b)$ and so $b=0$

So $b=0$ and :
$P(0,1)$ $\implies$ $b+ab=a+1$ and so $a=-1$
$P(-1,1)$ $\implies$ $a+bc=c-1$ and so $c=0$
The system above becomes then $f(x)=f(-x)$ and the function is even.

$P(\frac x2,\frac x2)$ $\implies$ $f(x)+f(\frac x2)^2=f(\frac {x^2}4)+\frac{x^2}2+1$

$P(\frac x2,-\frac x2)$ $\implies$ $-1+f(\frac x2)^2=f(\frac {x^2}4)-\frac{x^2}2+1$

Subtracting these two lines implies $\boxed{f(x)=x^2-1}$ which indeed is a solution

Let $P(x,y)$ be the assertion.
First see that no constant functions exist.
$P(x,0)\implies f(x)+f(x)f(0)=f(0)+1\implies (f(0)+1)(f(x)-1)=0$
As $f$ is non constant $\exists x$ such that $f(x)\neq 1\implies f(0)=-1$
$P(1,-1)\implies f(0)+f(1)f(-1)=f(-1)-1\implies f(-1)(f(1)-1)=0$
If $f(1)=1$
$P(x,1)\implies f(x+1)=2x+1\implies \boxed{f(x)=2x-1\text{ for all } x\in\mathbb{R}}$
If $f(-1)=0$.
$P(-1,-1)\implies f(-2)+f(-1)^2=f(1)+3\implies f(-2)=f(1)+3$.
$P(1,-2)\implies f(-1)+f(1)f(-2)=f(-2)-3\implies f(-2)(f(1)-1)=-3$
$\implies (f(1)+3)(f(1)-1)=-3$
$\implies f(1)^2+2f(1)=0\implies f(1)=0\text{ or } f(1)=-2$
$\text{ If }f(1)=0$
$P(x,1)\implies f(x+1)=f(x)+2x+1\implies f(x-1+1)=f(x-1)+2x-2+1$
$\implies f(x)=f(x-1)+2x-1$
$P(x,1)\implies f(x-1)=f(-x)-2x+1$
but we have just proved $f(x)=f(x-1)+2x-1$.
Comparing these two equations, we get $f(x)=f(-x)\text{ for all } x\in\mathbb{R}$
$P(x,-y)\implies f(x-y)+f(x)f(-y)=f(-x)-2xy+1\text{ but as }f(x)=f(-x)$
$f(x-y)=f(xy)-f(x)f(y)-2xy+1$
$P(x,y)\implies f(x+y)+f(x)f(y)=f(xy)+2xy+1$
Comparing these two equations, we get
$f(x+y)-f(x-y)=4xy$
Take $x=y$ to get $f(2x)=4x^2-1$,
so, $x\rightarrow \frac{x}{2}\implies \boxed{f(x)=x^2-1\text{ for all } x\in\mathbb{R}}\text{ is another solution}$
If $f(1)=-2$
$P(x,1)\implies f(x+1)=3f(x)+2x+1\implies f((x-1)+1)=3f(x-1)+2x-1\implies f(x)=3f(x-1)+2x-1$
$P(x,-1)\implies f(x-1)=f(-x)-2x+1$
$\implies 3(f(x-1))=3f(-x)-6x+3=f(x)-2x+1$
$\implies f(x)+4x=3f(-x)+2$
Now $x\rightarrow -x\implies f(-x)-4x=3f(x)+2$
Adding these two, we get, $f(x)+f(-x)=-2$
$P(x,-y)\implies f(x-y)+f(x)f(-y)=f(-xy)-2xy+1$
$\implies f(x-y)+f(x)\left[-2-f(y)\right]=-2-f(xy)-2xy+1$
$\implies f(x-y)-2f(x)+2xy+1=f(x)f(y)-f(xy)$
But $P(x,y)\implies f(x)f(y)-f(xy)=2xy+1-f(x+y)$
Therefore, $2xy+1-f(x+y)=f(x-y)-2f(x)+2xy+1$
$\implies f(x-y)+f(x+y)=2f(x)$
$x=y\implies -1+f(2x)=2f(x)\implies f(2x)-2f(x)=1$
We have already proved $f(x)=3f(x-1)+2x-1\stackrel{x=-1}{\implies} f(-2)=1$
As $f(x)+f(-x)=-2\implies f(-2)=3$
$P(x,2)\implies f(x+2)+f(x).f(2)=f(2x)+4x+1\implies f(2x)=f(x+2)-3f(x)-4x-1$
We have already proved
$f(x+1)=3f(x)+2x+1$
$\implies f((x+1)+1)=3f(x+1)+2x+3=9f(x)+6x+3+2x+3$
$\implies f(x+2)=9f(x)+8x+6$
Therefore $9f(x)+8x+6-3f(x)-4x-1=f(2x)$
But $f(2x)=1+2f(x)\implies 6f(x)+4x+5-2f(x)=1\implies 4f(x)=-4x-4$
$\implies \boxed{f(x)=-x-1\text{ for all }x\in\mathbb{R}}$

[hide="Solution"]Easy to get $f(0)=-1$ and $(f(1)-1)f(-1)=0$. So if $f(1)=1$, then $P(x-1,1)\implies f(x)=2x-1$ which is definitely a solution. So assume $f(-1)=0$. Let $f(1)=c$.

$P(-x+1,-1)\implies f(-x)-f(x-1)=2x-1$
$P(x-1,1)\implies f(x)+cf(x-1)-f(x-1)=2x-1$
Combining these we get $f(-x)=f(x)+cf(x-1)\quad (1)$
So $f(x)=f(-x)+cf(-x-1)$
Adding them gives $c[f(x-1)+f(-x-1)]=0$ So at least one of them is zero. Suppose $c=0$. Then (1) yields $f(x)=f(-x)$. Now subtracting $P(x,-x)$ from $P(x,x)$ gives $f(2x)=4x^2-1\implies f(x)=x^2-1$ which is another valid solution.

Now assume $f(x-1)+f(-x-1)=0\implies f(x)+f(-x-2)=0$. So $f(-2)=-f(0)=1$. And now $P(-1,-1)$ implies $f(1)=-2$.
$P(x,-1)\implies f(x-1)=f(-x)-2x+1$
$P(x,1)\implies f(x+1)=3f(x)+2x+1$
So $-f(x)=f(-x-2)=f(x+1)+2x+3=3f(x)+4x+2$ $\implies f(x)=-(x+1)$

So all the functions are $f(x)=2x-1,x^2-1,-x-1$[/hide]

[quote="Goblik"]Why $f(x)=1$ for all $x$ don't satisfy in this problem?[/quote]
Because with $f(x)=1$ $\forall x$ :

$LHS$ of functional equation is $2$
$RHS$ of functional equation is $2+2xy$

And obviously we dont have $LHS=RHS$ $\forall x,y$

I hate casework.
[hide=Solution]
The functions are $2x-1, x^2-1,$ and $-x-1$.

Let $P(x, y)$ be the assertion that $f(x+y) + f(x)f(y) = f(xy) + 2xy + 1$.
$P(0, 0)$ yields that $f(0)^2 = 1$, so $f(0) = 1$ or $f(0) = -1$. However, if $f(0) = 1$, then $P(x, 0)$ tells us that $f(x)$ is a constant function, which is clearly false. Thus, $f(0) = -1$. $P(1, -1)$ gives us $f(1)f(-1) = f(-1)$, so $f(1) = 1$ or $f(-1) = 0$.

[b]Case 1: $f(1) = 1$[/b]
$P(x - 1, 1)$ shows that $f(x) = 2x - 1$.

[b]Case 2:$f(-1) = 0$[/b]
$P(-2, 1)$ shows that $f(-2)f(1) = f(-2) - 3$ and $P(-1, -1)$ implies $f(-2) = f(1) + 3$. Putting together these two observations, we see that $f(1) = f(1)f(-2)$, so $f(1) = 0$ or $f(-2) = 1$.

[b]Subcase 1:$f(1) = 0$[/b]
$P(x - 1, 1)$ and $P(-x + 1, -1)$ tell us that $f(x) = f(-x)$. $P(-x, x)$ implies $-1 + f(-x)f(x) = f(-x^2) - 2x^2 + 1$, and substituting $f(x)$ for $f(-x)$ and $f(x^2)$ for $f(-x^2)$ gives us $f(x)^2 = f(x^2) - 2x^2 + 2$. $P(x, x)$ shows that $f(2x) + f(x)^2 = f(x^2) + 2x + 1$. Subtracting this equation from the previous one, we get $f(2x) = 4x^2 - 1$, or $f(x) = x^2 - 1$.

[b]Subcase 2: $f(-2) = 1$[/b]
We can easily show that $f(-2) = 1$ implies that $f(1) = -2$.
$P(-x, -1)$ gives us
$f(-x-1) = f(x) + 2x + 1$. (1)
 $P(x, 1)$ gives us
$f(x+1) = 3f(x) + 2x + 1$.      (2)
Subtracting, we get
$f(x+1) - f(-x-1) = 2f(x)$, or equivalently $f(x) - f(-x) = 2f(x-1)$. Substituting $x$ for $-x$, we can easily see that $f(x-1) = -f(-x-1)$. (3)
Substituting $x-1$ in (2), multiply (1) by 3, and subtracting those two equations, we get $f(-x-1) = 3f(x-1) + 4x$. Using (3), we immediately get $f(-x-1) = x$, or $f(x) = -x - 1$.

All those functions work, so QED.
[/hide]

This problem took a really long while for me... Well, here's my solution. Feedback would be greatly appreciated - it is loooong (4 o's to indicate it's pretty long, but not insane).
[hide="Spoilers!"]
We claim that the only possible functions $f$ are $f(x)=2x-1, f(x)=-x-1, f(x)=x^2-1$. It is easy to see that these values satisfy the problem condition.

Using $x=0, y\in\mathbb{R}$ is any real number, we have $f(y)(f(0)+1)=f(0)+1$.

If $f(0)\neq -1, f(y)=1$ for all real number $y$. Choosing $x=y=10$ gives us $1+1=1+200+1$, which is clearly not valid.
Therefore, $f(0)=-1$.

Using $x=1, y=-1$, we have $f(0)+f(1)f(-1)=f(-1)-2+1, f(-1)(f(1)-1)=0$. Therefore, $f(1)=1$ or $f(-1)=0$.

If $f(1)=1$, using $y=1$ gives us $f(x+1)+f(x)f(1)=f(x)+2x+1, f(x+1)=2x+1$ for all real $x$. Therefore, $f(x)=2x-1$ for all real $x$.

If $f(1)\neq 1$, $f(-1)=0$ and denote $c=f(1)$.
Using $y=1$, we have $f(x+1)+f(x)c=f(x)+2x+1$, or $f(x+1)=(1-c)f(x)+2x+1(\clubsuit)$ for all real $x$.
Using $y=-1$, we have  $f(x-1)+f(x)f(-1)=f(-x)-2x+1, f(x-1)=f(-x)-2x+1$ for all real $x$.
Therefore, $f(-x-1)=f(x)+2x+1(\star)$ for all real $x$.

Multiplying $(\star)$ by $1-c$ and subtract $(\clubsuit)$gives us $(1-c)f(-x-1)-f(x+1)=(2x+1)(-c)$ for any real $x$. Therefore, we also have $(1-c)f(x)-f(-x)=(2x-1)(-c)$, $(1-c)f(-x)-f(x)=(-2x-1)(-c)$ for all real $x$.

If $c\neq 0$, Solving for $f(x)$, we get that $f(x)=\frac{-2xc^2}{c^2-2c}-1$ for all real $x$. Using $x=1$, we get $c(c+2)(c-1)=0$. Since $c\neq 0,1$, we get $f(x)=-x-1$.

If $c=0$, then our equation gives us $f(x)=f(-x)$. Using $y=x$, we have $f(2x)+f(x)^2=f(x^2)+2x^2+1$. Furthermore, using $y=-x$, we have $f(0)+f(x)f(-x)=f(-x^2)-2x^2+1$.
Using $f(x)=f(-x)$ and subtracting, we have $f(2x)=4x^2-1$ for all real $x$. Therefore, $f(x)=x^2-1$.

Having exhausted all possible value of $f(1)=c$, the only possible functions $f$ are $f(x)=x^2-1, 2x-1, -x-1$, as desired.
[/hide]

[quote=IMO ShortList 2005 A4]Find all functions $ f: \mathbb{R}\to\mathbb{R}$ such that $ f(x+y)+f(x)f(y)=f(xy)+2xy+1$ for all real numbers $ x$ and $ y$.

[i]Proposed by B.J. Venkatachala, India[/i][/quote]

[hide=solution] Answer: The only functions which work are $f(x)=2x-1$ for all $x$, $f(x)=x^2-1$ for all $x$ and $f(x)=-(x+1)$ for all $x$. It is clear that they indeed satisfy the desired conditions.

Let $P(x,y)$ denote the given assertion and $f$ be a function which works. Let $h(x)=f(x)-x^2+1$ and $g(x)=f(x)+x+1$. I shall use them later on.

Note that $P(0,0) \Longrightarrow f(0)=\pm 1$ so if $P(0)=1$ then $P(x, 0) \Longrightarrow f(x)=1$ for all $x$ which is not a valid function. So $f(0)=-1$. Observe $$P(x, 1) \Longrightarrow f(x+1)+f(x)\cdot \left(f(1)-1\right)=2x+1$$ and $P(2, 2) \Longrightarrow f(2)=\pm 3$. Henceforth, $P(x, 1)$ is called the $(+)$. Plugging $x=1$ in $(+)$ yields $$f(1)(f(1)-1)=3-f(2) \in \{0, 6\} \Longrightarrow f(1) \in \{0, 1, -2, 3\}.$$ Consider the four cases for different values of $f(1)$ as follow.

Firstly, the easier ones.

[b]Case 1.[/b] $f(1)=1$

Our equation $(+)$ becomes $f(x+1)=2x+1 \Longrightarrow f(x)=2x-1$ for all reals $x$ which is one of the claimed solutions.

[b]Case 2.[/b] $f(1)=3$

Put $x=-1$ in $(+)$ to get $f(-1)\cdot \left(f(1)-1\right)=0 \Longrightarrow f(-1)=0$. Observe $P(-1, -1) \Longrightarrow f(-2)=f(1)+3 \Longrightarrow f(-2)=6$ but $x=-2$ in $(+)$ gives $f(-2)\cdot \left(f(1)-1\right)=-3 \Longrightarrow f(-2)=-1.5$ so we have a contradiction!

Harder cases incoming...

[b]Case 3.[/b] $f(1)=0$

Notice that $P(x, y)$ is equivalent to the assertion $$Q(x, y) \overset{\text{def}}{:=}h(xy)=h(x+y)+h(x)h(y)+(x^2-1)h(y)+(y^2-1)h(x). $$ Equation $(+)$ boils down to $f(1+x)-f(x)=2x+1 \Longrightarrow h(1+x)=h(x)$ for all reals $x$. Notice that $$Q(1+x, y)-Q(x, y) \Longrightarrow h(xy+y)-h(xy)=(2x+1)h(y).$$ Plugging $x=\frac{1}{y}$ for non zero $y$ in the last equation gives $\frac{h(y)}{y}=0$ so $h \equiv 0$ and we get $f(x)=x^2-1$ for all reals $x$, which is again one of the stated solutions.

[b]Case 4.[/b] $f(1)=-2$

Equation $(+)$ is equivalent to $g(1+x)=3g(x)$ and the assertion $P(x, y)$ is equivalent to $$R(x, y) \overset{\text{def}}{:=} g(x+y)+g(x)g(y)=g(xy)+(1+x)g(y)+(1+y)g(x).$$ Evidently, $$R(1+x, y)-3R(x, y) \Longrightarrow g(xy+y)-3g(xy)=(2x+1)g(y).$$ Plugging $x=\frac{1}{y}$ we conclude $2\left(\frac{1}{y}-1\right)g(y)=0$ so $g(y)=0$ for all $y \ne 1$, which combined with $g(1)=0$ yield $g \equiv 0$ or $f(x)=-(x+1)$ for all reals $x$, another one of the mentioned solutions.

By our rather long analysis, it is clear that these are the only valid functions.

End of Story.

[/hide]

[hide=Solution]
Answer: $f(x)=2x-1, f(x)=x^2-1,$ and $f(x)=-1-x$ are the three solutions.

Let $P(x, y)$ denote the assertion that \[f(x+y)+f(x)f(y)=f(xy)+2xy+1\] for all real numbers $x, y$. Then $P(0, 0)\implies f(0)^2=1\implies f(0)=1 \ \text{or} \ f(0)=-1$.

[b]Case 1[/b]: $f(0)=1$. Then $P(x, 0)\implies 2f(x)=2\implies f(x)=1$, which clearly doesn't satisfy the original FE.

[b]Case 2[/b]: $f(0)=-1$. Then \begin{align*}P(1, -1)&\implies -1+f(1)f(-1)=f(-1)-1 \\ &\implies f(-1)=0 \ \text{or} \ f(1)=1.\end{align*}
[list]
[*][b]Subcase 2.1[/b]: $f(1)=1$. Then \begin{align*}P(x, 1)&\implies f(x+1)+f(x)=f(x)+2x+1) \\ &\implies f(x+1)=2x+1 \\ &\implies f(x)=2x-1.\end{align*} We can easily check that $\boxed{f(x)=2x-1}$ indeed satisfies the original equation, as \[f(x+y)+f(x)f(y)=2x+2y-1+4xy-2x-2y+1=f(xy)+2xy+1.\]
[*][b]Subcase 2.2[/b]: $f(-1)=0$. Let $g(x)=f(x)+1$, and we have that $P(x, y)$ is equivalent to the assertion \[Q(x, y):=g(x+y)+g(x)g(y)-g(x)-g(y)=g(xy)+2xy\] for all reals $x, y$. Note that $g(0)=0$ and $g(-1)=1$.

$Q(-1, -1)$ gives $g(-2)=g(1)+3$, and $Q(1, -2)$ gives \[1+g(1)g(-2)-g(1)-g(-2)=g(-2)-4 \implies g(1)=1 \ \text{or} \ -1\] after substituting $g(-2)=g(1)+3$.
[list]
[*][b]Subcase 2.2.1[/b]: $g(1)=1$. Then $Q(x, 1)\implies g(x+1)-1=g(x)+2x$. We also have that $Q(-x, -1)\implies g(-x-1)-1=g(x)+2x$, so $g(x+1)=g(-(x+1))$ for all $x$, and $g$ is even. But \begin{align*}Q(x, -x)&\implies g(x^2)-2g(x)=g(x^2)-2x^2 \\ &\implies g(x)=x^2.\end{align*} Because $f(x)=g(x)-1$, the $g(x)=x^2$ corresponds to $\boxed{f(x)=x^2-1}$ for all real $x$. This can indeed be checked to satisfy the original FE, as \[f(x+y)+f(x)f(y)=x^2+2xy+y^2+(xy)^2-x^2-y^2+1=f(xy)+2xy+1.\]
[*][b]Subcase 2.2.2[/b]: $g(1)=-1$. Then $Q(x, 1)\implies g(x+1)=3g(x)+2x-1 \ \ \ (*)$. Plugging $x=-\tfrac{1}{2}$ into $(*)$ gives \[g\left(\tfrac{1}{2}\right)=3g\left(-\tfrac{1}{2}\right)-2. \ \ \ (1)\] Replacing $x$ with $2x-\tfrac{1}{2}$ in $(*)$ gives \[g\left(2x+\tfrac{1}{2}\right)=3g\left(2x-\tfrac{1}{2}\right)+4x-2. \ \ \ (2)\] Write \begin{eqnarray*}Q\left(2x, -\tfrac{1}{2}\right)\implies g\left(2x-\tfrac{1}{2}\right)+g(2x)g\left(-\tfrac{1}{2}\right)-g(2x)-g\left(-\tfrac{1}{2}\right)=g(-x)-2x\end{eqnarray*} and then use $(1), (2)$ to get \begin{align*}Q\left(2x, \tfrac{1}{2}\right)&\implies g\left(2x+\tfrac{1}{2}\right)+g(2x)g\left(\tfrac{1}{2}\right)-g(2x)-g\left(\tfrac{1}{2}\right)=g(x)+2x \\ &\implies 3\left[g\left(2x-\tfrac{1}{2}\right)+g(2x)g\left(-\tfrac{1}{2}\right)-g(2x)-g\left(-\tfrac{1}{2}\right)\right]=g(x)-2x \\ &\implies 3\left[g(-x)-2x\right] = g(x)-2x\ \\ &\implies 3g(-x) = g(x) + 4x. \ \ \ (3)\end{align*} Replacing $x$ with $-x$ in $3g(-x)=g(x)+4x$ yields \[3g(x)=g(-x)-4x, \ \ \ (4)\] and solving for $g(x)$ in $(3)$ and $(4)$ returns $g(x)=-x$. The corresponding function in $f$ for $g(x)=-x$ is $\boxed{f(x)=-1-x}$, and it is easy to check that it works: \[f(x+y)+f(x)f(y)=-1-x-y+xy+x+y+1=f(xy)+2xy+1.\]

[/list]
[/list]
---------
We are done, having exhausted all cases.
[/hide]

Let $P(x,y)$ denote the assertion that $ f(x+y)+f(x)f(y)=f(xy)+2xy+1$.
$$P(x,0) \implies (1+f(0))(f(x))=1+f(0) \implies f(0)=-1$$
If $f(1)=1$, then $P(x-1,1) \implies f(x)=2x-1$. Now suppose that $f(1) \neq 1$ and let $c=f(1)$.
$$P(1,-1) \implies f(1)f(-1)=f(-1) \implies f(-1)=0$$
$$P(x,-1) \implies f(x-1)=f(-x)-2x+1$$
$$P(x-1,1) \implies f(x)+cf(x-1)=f(x-1)+2x-1 \implies f(x)+c(f(-x)-2x+1)=f(-x) \implies (c-1)f(-x)=2cx-c-f(x) \ \ (Q(x))$$
$$Q(-1) \implies c(c-1)=-3c \implies c=0 \  \text{or} \ c=-2$$
Let's check the case $c=-2$.
$$(c-1)Q(-x)-Q(x) \implies (c-1)f(-x)=-(c-1)(2cx+c+(c-1)f(x))=2cx-c-f(x) \implies 3(-4x-2-3f(x))=-4x+2-f(x) \implies f(x)=-x-1$$
Now let's check the case when $c=0$. Note that $Q(x) \implies f(x)=f(-x)$
$$P(x,x)-P(x,-x) \implies f(2x)+1=4x^2 \implies f(x)=x^2-1$$
To conclude, the only solutions are $f(x)=x^2-1$ , $f(x)=-x-1$ and $f(x)=2x-1$.

1. y=1
2. y=y+1 (expand)
3. use symmetry
Done. That easy (try it)

Beautiful FE. Worthy of its position as A4. Here's my solution (probably different):
We claim that the only solutions that work are $f(x)=2x-1,f(x)=x^2-1,f(x)=-x-1 \text{ } \forall x \in \mathbb{R}$. One can easily check that all three of these satisfy the given equation. We'll now show that these are the only functions that work.

Let $P(x,y)$ denote the given assertion. Note that no constant function satisfies the given equation. So from now on assume that $f$ is non-constant.

$$P(x,0) \Rightarrow f(x)+f(x)f(0)=f(0)+1 \Rightarrow (f(x)-1)(f(0)+1)=0 \Rightarrow f(0)=-1$$

$$P(1,-1) \Rightarrow f(1)f(-1)=f(-1) \Rightarrow f(1)=1 \text{ OR } f(-1)=0$$

[b]CASE-1[/b] $(f(1)=1):$ $$P(x-1,1) \Rightarrow f(x)+f(x-1)=f(x-1)+2(x-1)+1 \Rightarrow \boxed{f(x)=2x-1}$$

[b]CASE-2[/b] $(f(-1)=0):$ Let $f(1)=k \neq 1$. Then $P(-1,-1) \Rightarrow f(-2)=f(1)+2+1=k+3$. Also, $P(2,2) \Rightarrow f(2)^2=9 \Rightarrow f(2)= \pm 3$

[b]SUBCASE-2.1[/b] $(f(2)=3):$ Note that $P(2,-2) \Rightarrow -1+f(2)f(-2)=f(-4)-7 \Rightarrow f(-4)=3(k+3)+6=3k+15$

And, $P(2,1) \Rightarrow f(3)+f(2)f(1)=f(2)+4+1 \Rightarrow f(3)=3(1-k)+5=8-3k$

But, $P(-3,-1) \Rightarrow f(-4)=f(3)+6+1 \Rightarrow 3k+15=8-3k+7 \Rightarrow k=0 \Rightarrow f(1)=0$

Now, $P(x-1,1) \Rightarrow f(x)=f(x-1)+2(x-1)+1$ and $P(1-x,-1) \Rightarrow f(-x)=f(x-1)+2(x-1)+1$

Together, These two statements give that $f$ is an even function. Then $$P(x,x)-P(x,-x) \Rightarrow f(2x)-f(0)=2x^2-(-2x^2) \Rightarrow f(2x)=4x^2-1 \Rightarrow \boxed{f(x)=x^2-1}$$

[b]SUBCASE-2.2[/b] $(f(2)=-3):$ Note that $P(2,-2) \Rightarrow -1+f(2)f(-2)=f(-4)-7 \Rightarrow f(-4)=-3(k+3)+6=-3k-3$

And, $P(2,1) \Rightarrow f(3)+f(2)f(1)=f(2)+4+1 \Rightarrow f(3)=-3(1-k)+5=3k+2$

But, $P(-3,-1) \Rightarrow f(-4)=f(3)+6+1 \Rightarrow -3k-3=3k+2+7 \Rightarrow 6k=-12 \Rightarrow k=-2 \Rightarrow f(1)=-2$

Now, $P(x,1) \Rightarrow f(x+1)=3f(x)+2x+1$ and $P(-x,-1) \Rightarrow f(-x-1)=f(x)+2x+1$

Together these two statements give that $f(x+1)=2f(x)+f(-x-1)$. Putting $x \rightarrow x-1$, we have $f(x)-f(-x)=2f(x-1)$

Also, $P(-x,1) \Rightarrow f(1-x)=3f(-x)-2x+1=3(f(x)-2f(x-1))-2x+1$. Putting $x \rightarrow x+1$ in this equation, we get that $$f(-x)=3(f(x+1)-2f(x))-2x-1=3f(x+1)-6f(x)-2x-1$$

And, $$P(x,-1) \Rightarrow f(x-1)=f(-x)-2x+1 =(3f(x+1)-6f(x)-2x-1)-2x+1=3(3f(x)+2x+1)-6f(x)-4x=3f(x)+2x+3$$

Thus, $$P(x-1,1) \Rightarrow f(x)=3f(x-1)+2x-1 \Rightarrow f(x-1)=3(3f(x-1)+2x-1)+2x+3=9f(x-1)+8x \Rightarrow f(x-1)=-x$$

Putting $x \rightarrow x+1$, we get our final solution as $\boxed{f(x)=-x-1}$

Found the source of the problem I was looking for :D

Let $P(x,y)$ be the assertion of the problem statement. $P(0,0)$ gives $f(0)^2 = 1$ or $f(0)=-1$ or $1$.

If $f(0) = 1$, $P(x,0)$ gives $2f(x) = 2$ or $f(x) = 1$ for all $x$ which is not a solution. Thus $f(0) = -1$.

If $f(1) = 1$, then $P(x,1)$ gives $f(x+1) +f(x) = f(x) + 2x+1\implies \boxed{f(x) = 2x-1}$ which is a solution.

Now assume $f(1) = c\neq 1$. $P(-1,1)$ gives $f(0) + f(1)f(-1) = f(-1) - 1 \implies f(-1)(f(1) - 1)=0$ or $f(-1) = 0$. Now $P(x-1,1)$ gives $f(x) + cf(x-1) = f(x-1) + 2x-1$. Also $P(x,-1)$ gives $f(x-1) = f(-x) - 2x+1$.

Thus
\begin{align*}
f(x) &= (1-c)f(x-1) + 2x-1\\
& = (1-c)(f(-x)-2x+1) + 2x-1\\
& = f(-x)-2x+1-cf(-x)+c(2x-1) + 2x+1\\
& = f(-x) - cf(-x) + c(2x-1) = (1-c)f(-x) + c(2x-1).
\end{align*}

Plugging $x=-1$ in the above equation gives $f(-1) = f(1)(1-c)-3c\implies c(1-c) = 3c\implies c=0,-2$.

If $c=0$ then \[f(x) = (1-c)f(-x)+c(2x-1) = f(-x).\] We have $P(x,-x)\implies -1 + f(x)^2 = f(x^2) - 2x^2 + 1$ and $P(x,x)$ gives $f(2x) + f(x)^2 = f(x^2) +2x^2 + 1$ thus \[f(2x) + 1 =  4x^2\implies \boxed{f(x) = x^2-1}.\]

If $c=-2$ then \[f(x) = (1-c)f(-x)+c(2x-1) = 3f(-x) - 2(2x-1).\] Solving linear equation by plugging $x=-x$ in the previous equation gives \[-8f(x) = -4x+2 + 12x+ 6 = 8x+8\implies \boxed{f(x) = -(x+1)}.\]

Taking $y=0$ gives $f(x)(1+f(0))=f(0)+1$. Clearly $f(x)=1$ is a not a solution, so we must have $f(0)=-1$. Applying $(x,y)=(2,2)$ yields $f(2)=\pm 3$. Taking $(x,y)=(1,1)$ gives $f(2)+f(1)^2=f(1)+3$.

[b]Case 1: f(2)=3[/b] Then $f(1)^2=f(1)$, so $f(1)=1$ or $f(1)=0$. Suppose $f(1)=1$. Plugging $y=1$,
\begin{align}
f(x+1)+f(x)f(1)&=f(x)+2x+1
\end{align}
Hence, $f(x+1)=2x+1$. Thus, $\boxed{f(x)=2x-1}$ for all real $x$, and it is easy to verify this works. Now we visit the case when $f(1)=0$. Note $(1)$ transforms to $$f(x+1)=f(x)+2x+1\implies f(x+1)-(x+1)^2=f(x)-x^2.$$
A simple double induction proves that $f(x+n)=f(x)+2nx+n^2$ for all integers $n$. Another induction shows that $f(n)=n^2-1$. Taking $y=n$ for any integer $n$, we see $$f(nx)=f(x+n)+f(x)f(n)-2nx-1 = f(x)+2nx+n^2+(n^2-1)f(x)-2nx-1=n^2f(x)+n^2-1.$$
Note that taking $n=-1$ gives $f(-x)=f(x)$. Plugging $y=x$ gives
\begin{align*}\tag{2}
f(x^2)+2x^2+1=f(2x)+f(x)^2=f(x)^2+4f(x)+3
\end{align*}
Now taking $y=-x$ grants \begin{align*}\tag{3} f(0)+f(x)f(-x)=f(-x^2)-2x^2+1\implies f(x)^2=f(x^2)-2x^2+2\end{align*}
Combining $(2)$ and $(3)$ gives $2x^2+1-(4f(x)+3)=-2x^2+2$. Therefore $\boxed{f(x)=x^2-1}$.

[b]f(2)=-3[/b] In this case $f(1)^2-f(1)-6=0$, so that $f(1)=3$ or $f(1)=-2$. Suppose $f(1)=3$. Trying $y=1$ grants $$f(x+1)=-2f(x)+2x+1.$$ Computing, we see $f(3)=11$, $f(5)=39$ and $f(6)=-67$. Now taking $(x,y)=(2,3)$ gives $$f(5)+f(2)f(3)=f(6)+13\implies f(6)=-7,$$ which does not match with $f(6)=-67$. Thus, we must have $f(1)=-2$. Again, we take $y=1$ to get $$f(x+1)=3f(x)+2x+1.$$ A little bashing gives $f(x-1)=(f(x)-2x+1)/3$. Note $(x,-1)$ grants $$f(-x)=2x-1+f(x-1)=2x-1+(f(x)-2x+1)/3.$$
Replacing $x$ with $-x$ in the above equation and adding gives $f(x)+f(-x)=-2$. Plugging $f(-x)=-2-f(x)$ back into the equation gives $$-2-f(x)=2x-1+(f(x)-2x+1)/3.$$ Solving, we get $\boxed{f(x)=-x-1}$. We can easily verify all the derived solutions work.

Let $Q(x)$ be the assertion given in the problem statement. Then, $Q(x, 0)$ gives:
$f(x)(f(0) + 1) = f(0) + 1$. As $f \equiv 1$ is not a solution, we have $f(0) = -1$.
Now, let $g(x) = f(x) + 1$.
Then $Q(x)$ can be written as $P(x)$:
$g(x+y) - (g(x) + g(y)) = g(xy) - g(x)g(y) + 2xy$. We have $g(0) = 0$.
$P(1, -1)$ gives: $g(-1) = 1$ or $g(1) = 2$
Case 1: $g(1) = 2$
$P(x,1)$ gives $g(x+1) = 2(x+1)$ so $g(x) \equiv 2x$. One can see that this is a valid solution.
Case 2: $g(-1) = 1$
$P(x, -1)$ gives $g(x-1) - 1 = g(-x) - 2x$. Call this assertion $R(x)$.
$R(2)$ gives $g(-2) = g(1) + 3$
Considering $P(-2, 1)$ with the above relation and solving for $g(1)$ gives $g(1) = 1$ or $-1$.
Subcase 2.1: $g(1) = 1$
$P(x-1, 1)$ gives $g(x-1) = g(x) - 2x + 1$, which combines with $R(x)$ to give us that $g$ is even.
Using this knowledge and considering $P(x, x)$ and $P(x, -x)$ together gives $g(2x) = 4x^2$, thus $g(x) \equiv x^2$, which can be checked to be a valid solution.
Subcase 2.2: $g(1) = -1$
$P(x-1, 1)$ gives $g(x) = 3g(x - 1) + 2x - 3$ -> $(1)$
Combining this with $R(x)$ gives $g(x) = 3g(-x) - 4x$
Considering $R(-x)$ gives $3g(x+1) + 4x + 3 = g(-x-1) - 1 = g(x) + 2x$ -> $(2)$
Solving for $g(x)$ using $(1)$ and $(2)$ gives $g(x) = -x$, thus $g(x) \equiv -x$, which can be checked to be a valid solution.

Thus, the solution set is $f(x) \equiv 2x - 1$, $f(x) \equiv -x - 1$, and $f(x) \equiv x^2 - 1$.

Probably the same as some other solutions in the thread.

First $(0,0)$ yields $f(0)^2=1\implies f(0)=\pm 1$.

[b]Case 1:[/b] $f(0)=1$. Then $y=0$ yields $f(x)=1$, which doesn't work.

[b]Case 2:[/b] $f(0)=-1$. Now $(1,-1)$ yields $f(1)f(-1)=f(-1)$ so $f(-1)=0$ or $f(1)=1$.

[i]Case 2a:[/i] $f(1)=1$. Then $y=1$ gives $f(x+1)=2x+1$ so $\boxed{f(x)=2x-1}$ here.

[i]Case 2b:[/i] $f(-1)=0$ and $f(1)\neq 1$. First $(2,2)$ yields $f(2)^2=9\implies f(2)=\pm 3$.

[u]Case 2ba:[/u] $f(2)=-3$. Now $(1,1)$ yields $f(1)^2-f(1)=6$ so $f(1)$ is $3$ or $-2$. Note from $y=-1$ that $f(x-1) =f(-x)-2x+1$, so letting $x=3,2$ gives $f(-3)=2$ and that $f(-2)$ is either $6$ or $1$ depending on our choice of $f(1)$. But now $(-3,1)$ yields $f(-2)+2f(1)=-3$, from which we find that $f(-2)=1$ and $f(1)=-2$. Now $y=-1$ yields $f(x-1)=f(-x)-2x+1$ and $y=1$ yields $f(x+1) =3f(x)+2x+1$. It follows that $f(-x)-2x+1=f(x-1) = \dfrac{f(x)-2x+1}{3}$, which rearranges as $3f(-x)+2=f(x)+4x$.This also implies $3f(x)+2=f(-x)-4x$, and now we solve to find $\boxed{f(x)=-1-x}$ for all $x$.

[u]Case 2bb:[/u] $f(2)=3$. Now $(1,1)$ yields $f(1)^2=f(1)$ so $f(1)\neq 1$ implies $f(1)=0$. Now $y=1$ gives $f(x+1)=f(x)+2x+1$ while $y=-1$ gives $f(x-1)=f(-x)-2x+1$. This last equation implies $f(-x-1)=f(x)+2x+1=f(x+1)$, so $f$ must be even. Now $y\to -y$ gives $f(x-y)+f(x)f(y)=f(xy)-2xy+1$, so subtracting from the original equation gives $f(x+y)-f(x-y) = 4xy$. It follows that $f(x+y)-(x+y)^2 = f(x-y) - (x-y)^2$ so $f(u)-u^2$ is constant, hence $u=0$ implies the constant is $-1$ and $\boxed{f(x)=x^2-1}$.

All cases are done so we have $\boxed{f(x) = x^2-1, -1-x, 2x-1}$.

Arghh, why so long BJV...
Again, I'll to make my solution motivated, but this becomes hard to do with FEs, so... well, here's my solution-

First we get the basics substitutions over with-

$P(x,0): f(x)(1+f(0))=(1+f(0))$. This immediately gives us that if $f(0) \neq -1 => f \equiv 1$ which of course doesn't work.
Hence $f(0)=-1$.

So we've just gotten some information about $f(0)$- the next natural substitution seems to be $P(x,-x)$:
$$P(x,-x): -1+f(x)f(-x)=f(-x^2)+1-2x^2$$
Putting in $x=1 =>-1+f(1)f(-1)=f(-1)-1=>f(-1)(f(1)-1)=0$.

Case 1: $f(1)=1$
$P(x,1): f(x+1)+f(x)f(1)=f(x)+2x+1=>f(x+1)=2x+1=>f(x)=2x-1$.

Ok, easiest solution found- time to deal with the other, rather more messier case:

Case 2: $f(-1)=0$
Let's try and extract as much as we can from such a nice zero:
$$P(x,-1): f(x-1)=f(-x)-2x+1=>f(-x-1)=f(x)+2x+1$$
$$P(x,-x-1): f(x)f(-x-1)=f(-x(x+1))-2x(x+1)+1=>f(x)^2+2xf(x)+f(x)=f(-x(x+1))-2x^2-2x+1$$
Putting in $x=1$, we get $f(1)^2+3f(1)=f(-2)-3$. But by the first equation, putting in $x=2 =>f(1)=f(-2)-3$.

Solving for $f(-2)$, we get $(f(2)-3)^2(f(-2)-1)=0=>f(-2)=3$ or $f(-2)=1$.

Sub-case 1: $f(-2)=1 =>f(1)=f(-2)-3=>f(1)=-2$.
Now the deal with both these cases is to play with $f(x+1),f(-x)$ and similar $f$s and see if we can get something...
As we've already gotten $f(-x-1)=f(x)+2x+1$, and $P(x,1)=>f(x+1)=3f(x)+2x+1$
$$=> f(-x)=f(x)+2f(x-1)=3f(x)+4x+2$$
Now $P(x-1,1): f(x)-2f(x-1)=f(x-1)+2x-1=>f(x)=3f(x-1)+2x-1=>f(x-1)=\frac{f(x)+1-2x}3$
$=>f(-x)-2x+1=\frac{f(x)+1-2x}3 =>f(-x)=\frac{4x-2+f(x)}3$.

Equating both expressions for $f(-x)$, we get $(3f(x)+4x+2) \cdot 3=4x-2+f(x)=>8f(x)=-8x-8 =>f(x)=-x-1$.
Aha, only one case left- we're almost there.

Subcase 2: $f(-2)=3 =>f(1)=\frac{f(-2)-3}3=>f(1)=0$
Okay we have another nice zero here- clearly it would be ideal to put in $(x,1)$ and see what we get:
$$P(x,1): f(x+1)=f(x)+2x+1$$
Wait this expression seems familiar- turning our pages back (or scrolling up :P ) we see that $f(x)+2x+1$ is also equal to $f(-x-1)$- hence $f$ is even.
Now the natural substitution is $P(x,-x): -1+f(x)^2=f(-x^2)+1-2x^2=>f(x^2)=f(x)^2+2x^2-2$.

But $P(x,x)$ gives us $f(2x)+f(x)^2=f(x^2)+2x^2+1=f(x)^2+2x^2-2+2x^2+1=>f(2x)=4x^2-1$ (I was so surprised this worked, but it did, so lucky me :) )

Thus our 3 solutions are $ f \equiv 2x-1, f \equiv -x-1$ and $f \equiv x^2-1$, and as all 3 work, we're finally done!

Let $P(x,y)$ denote the assertion.
-----
[b]Claim 1.[/b]     If $f(x)=2x-1$ for all $x$, then $f(0)=-1$ and $f(-1)=0$.

[i]Proof.[/i]     $P(0,0)$ immediately gives $f(0)=\pm1$, but if $f(0)=1$, then $P(x,0)$ gives $f\equiv1$, contradiction. Thus $f(0)=-1$.

    Now $P(1,-1)$ gives us $f(1)f(-1)=f(-1)$, so either $f(-1)=1$ or $f(-1)=0$. Say $f(-1)=1$. Then $P(x-1,1)$ immediately yields $f(x)=2x-1$, as claimed. $\blacksquare$
-----
[b]Claim 2.[/b]     The tuple $(f(-2),f(1),f(2))$ is either $(1,-2,-3)$ or $(3,0,3)$.

[i]Proof.[/i]     Note that $P(1,-2)$ gives $f(1)f(-2)=f(-2)-3$ and $P(2,-1)$ gives $f(1)=f(-2)-3$. Solving, $(f(-2),f(1))$ is either $(1,-2)$ or $(3,0)$.

    However $P(1,1)$ gives $f(2)=-f(1)^2+f(1)+3$, which equals $-3$ and $3$ respectively. $\blacksquare$
-----
[b]Claim 3.[/b]     If $(f(-2),f(1),f(2))=(1,-2,-3)$ then $f(x)=-x-1$ for all $x$.

[i]Proof.[/i]     Note that $P(1-x,-1)$ and $P(1+x,-1)$ give the two equations
    \begin{align*}
        f(-x)&=f(x-1)+2x-1,\\
        f(-x-1)&=f(x)+2x+1.
    \end{align*}
    respectively. Furthermore $P(x-1,1)$ and $P(-x-1,1)$ give the two equations
    \begin{align*}
        f(x)=3f(x-1)+2x-1,\\
        f(-x)=3f(-x-1)-2x-1.
    \end{align*}
    Solving the system of equations, we find $f(x)=-x-1$, as desired. $\blacksquare$
-----
[b]Claim 4.[/b]     If $(f(-2),f(1),f(2))=(3,0,3)$ then $f(x)=x^2-1$ for all $x$.

[i]Proof.[/i]     First $P(x-1,1)$ and $P(-x+1,-1)$ give \[f(x)=f(x-1)+2x-1=f(-x),\]
    so $f$ is even. With this, $P(x/2,x/2)$ and $P(x/2,-x/2)$ give the two equations
    \begin{align*}
        f(x)+f(x/2)^2&=f(x^2/4)+x^2/2+1,\\
        -1+f(x/2)^2&=f(x^2/4)-x^2/2+1.
    \end{align*}
    Subtracting, $f(x)=x^2-1$, as desired. $\blacksquare$
-----
This exhausts all possible cases, so we conclude the only answers are $f(x)=2x-1$, $f(x)=-x-1$, $f(x)=x^2-1$, which clearly work.

The solutions are $f(x)\equiv\boxed{2x-1,-x-1,x^2-1}$, which can all be checked to work. Now let $P(x,y)$ be the given FE, and suppose $f$ is a solution. Note that \[P(x,0)\implies f(x)[1+f(0)]=f(0)+1,\] and since $f\equiv 1$ isn't a solution, we have $\boxed{f(0)=-1}$. Now, we have \[P(1,-1)\implies -1+f(1)f(-1)=f(-1)-1\implies f(1)f(-1)=f(-1).\] If $f(-1)\ne 0$, then we have $f(1)=1$, so \[P(x,1)\implies f(x+1)+f(x)=f(x)+2x+1,\] so $f(x)\equiv 2x-1$, which is a solution. Thus, we may now assume $\boxed{f(-1)=0}$ and $f(1)\ne 1$.

We see that \[P(x,1)\implies \boxed{f(x+1)=2x+1+f(x)[1-f(1)]}\] and \[P(x+1,-1)\implies f(x)=f(-(x+1))-2x-1.\] Adding the two equations, we get \[\boxed{f(x+1)-f(-(x+1))=-f(1)f(x)}.\] Call the first boxed equation $Q(x)$ and the second one $R(x)$.

We have \[Q(-2)\implies 0=-3+f(-2)[1-f(1)]\implies f(-2)=\frac{3}{1-f(1)},\] and \[Q(1)\implies f(2)=3+f(1)[1-f(1)].\] Thus, \[R(1)\implies 3+f(1)[1-f(1)]-\frac{3}{1-f(1)}=-f(1)^2,\] so solving for $f(1)$ gives either $f(1)=0$ or $f(1)=-2$. We thus have two cases.

[b]Case 1[/b]: Suppose $f(1)=0$. Then, $Q(x)$ reduces to \[f(x+1)-f(x)=2x+1.\] It is not hard to show by induction that \[f(x+n)-f(x)=2nx+n^2\] for any integer $n$. Combined with $f(0)=-1$, we see that $f(n)=n^2-1$ for any integer $n$. Thus, \[P(x,n)\implies f(x)+2nx+n^2+f(x)(n^2-1)=f(nx)+2nx+1,\] so \[f(nx)=n^2f(x)+(n^2-1).\] We now have \begin{align*} P(x,x) &\implies f(2x)+f(x)^2=f(x^2)+2x^2+1 \\ &\implies 3+4f(x)+f(x)^2=f(x^2)+2x^2+1 \end{align*} and \begin{align*} P(x,2x) &\implies f(3x)+f(x)f(2x)=f(2x^2)+4x^2+1 \\ &\implies 9f(x)+8+f(x)[4f(x)+3]=4f(x^2)+3+4x^2+1 \\ &\implies 4f(x)^2+12f(x)+4=4f(x^2)+4x^2 \\ &\implies 3f(x)+f(x)^2+1=f(x^2)+x^2. \end{align*} Comparing the value for $f(x^2)$ we get from these two equations, we learn \[f(x)^2+4f(x)+2-2x^2 = f(x)^2+3f(x)+1-x^2,\] so $f(x)=x^2-1$, as desired. This completes Case 1.

[b]Case 2[/b]: Suppose $f(1)=-2$. Then, $Q(x)$ gives \[f(x+1)=2x+1+3f(x).\] It is not hard to show by induction that \[f(x+n)=3^nf(x)+(3^n-1)x+(3^n-n-1)\] for any integer $n\ge 0$. In particular, $f(n)=-n-1$ for integers $n\ge 0$. Thus, \[P(x,n)\implies 3^nf(x)+(3^n-1)x+(3^n-n-1)-(n+1)f(x)=f(nx)+2nx+1,\] so \[f(nx)=(3^n-n-1)f(x)+(3^n-1-2n)x+(3^n-n-2).\] We now have \begin{align*} P(x,x) &\implies f(2x)+f(x)^2=f(x^2)+2x^2+1 \\ &\implies 6f(x)+4x+5+f(x)^2=f(x^2)+2x^2+1 \\ &\implies f(x^2)=f(x)^2+6f(x)-2x^2+4x+4 \end{align*} and \begin{align*} P(x,2x) &\implies f(3x)+f(x)f(2x)=f(2x^2)+4x^2+1 \\ &\implies 23f(x)+20x+22+f(x)[6f(x)+4x+5]=6f(x^2)+4x^2+5+4x^2+1. \end{align*} Substituting the value for $f(x^2)$ from $P(x,x)$ into the above equation and simplifying, we get \[(x-2)f(x)=(2-x)(x+1),\] so $f(x)=-x-1$ when $x\ne 2$. However, $f(2)=-3$ as $2$ is an integer, so we have $f(x)=-x-1$, as desired.

This completes the proof.

Let $P(x,y) $ denotes the assertion.
If $a=0$ then $f(b)+f (0)f(b)=f(b)=f(0)+1\implies f(b)=-1, f(0)\ne -1\quad\forall b\in\mathbb R$
Similarly, if $b=0$ we obtain $f(a)=f(b)=-1, f(a)=-1\quad\forall a,b\in\mathbb R$

If $a=1, P(x-1,1)\implies f(x)=2x-1, a=-2\implies f(x)=-x-1\forall x\ne (0,1), f(x)=-x-1\forall x$

[hide=Correction in #24][quote="TheDarkPrince"]
\begin{align*}
f(x) &= (1-c)f(x-1) + 2x-1\\
& = (1-c)(f(-x)-2x+1) + 2x-1\\
& = f(-x)-2x+1-cf(-x)+c(2x-1) + 2x+1\\
& = f(-x) - cf(-x) + c(2x-1) = (1-c)f(-x) + c(2x-1).
\end{align*}
[/quote]

Nice solution TDP, just make it $2x-1$ in the third line :)[/hide]

The solutions are $f\equiv 2x-1$, $f\equiv x^2-1$, and $f\equiv -x-1$. It is easy to check that each works.

Set $x=y=0$ to yield $f(0)^2=1$. Set $y=0$ to yield $f(x)(f(0)+1)=f(x)+f(x)f(0)=f(0)+1$. Thus either $f(0)=-1$ or $f\equiv1$, which is clearly absurd. Then $f(0)=-1$. Let $y=-x$ to yield
\[-1+f(x)f(-x)=f(-x^2)-2x^2+1.\]
In particular, for $x=1$ this yields $f(1)f(-1)=f(-1)$, so either $f(1)=1$ or $f(-1)=0$. In the case of $f(1)=1$, we have $f(x+1)=f(x)=f(x)+2x+1$ so $f(x+1)=2[x+1]-1$, yielding the solution $f\equiv 2x-1$.

Now suppose $f(-1)=0$. Then we write the equation $f(-x-1)=f(x)+2x+1$. Take $y=-x-1$ so
\[f(-x^2-x)-2x^2-2x+1=f(x)f(-1-x)=f(x)f(x)+2xf(x)+f(x)\qquad (\diamondsuit).\]
So for $x=1$, this reads $f(-2)+1=f(1)^2+3f(1)$. But we also have $f(-2)=f(1)+3$, so
\[f(1)=f(1)^2+3f(1)\iff f(1)\in \{0,-2\}.\]
The first case yields the equation $f(x+1)=f(x)+2x+1$ so $f$ is even, so then $f(2x)+f(x)f(-x)=f(-x^2)+2x^2+1$, which implies $f(2x)=4x^2-1$ and $f\equiv x^2-1$.

Now we have $f(1)=-2$, $f(-1)=0$, and $f(0)=-1$. Take $x=1$ to yield $f(1+y)=3f(y)+2y+1$. Now, write
\[f(y)=3f(y-1)+2y-1=3[f(-y)-2y+1]+2y-1 = 3f(-y)-4y+2 = 3[3f(y)+4y+2]-4y+2=9f(y)+8y+8\implies f\equiv -y-1.\]

Let $P(x,y)$ be the assertion $f(x+y)+f(x)f(y)=f(xy)+2xy+1$.

$$P(x,0)\implies\left(1+f(0)\right)\left(f(x)-1\right)=0$$
If $f(0)\ne-1$ then $f(x)=1$, which doesn't work. Therefore, we must have $f(0)=-1$.

Also note that:
$$P(1,-1)\implies f(1)f(-1)=f(-1)$$
So either $f(1)=1$ or $f(-1)=0$.

[b]Case 1:[/b] $f(1)=1$

We have that:
$P(x-1,1)\implies\boxed{f(x)=2x-1}$, which works.

[b]Case 2:[/b] $f(-1)=0$

\begin{align*}
P(-x,-1)&\implies f(-x-1)=f(x)-2x+1\qquad(1) \\
P(x,1)&\implies f(x+1)+f(x)f(1)=f(x)+2x+1\qquad(2) \\
&\implies f(-x-1)+4x=f(x+1)+f(x)f(1)\qquad(3)
\end{align*}

Taking $x\mapsto-x-1$ in $(3)$, we get:
$f(x)=f(-x)+f(-x-1)f(1)-4x$

Substituting $(1)$, we can get an expression in terms of just $f(x)$, $f(-x)$, $x$, and constants.
$f(x)=f(-x)+f(x)f(1)+(2f(1)-4)x+f(1)$
$\implies f(x)\left(1-f(1)\right)=f(-x)+(2f(1)-4)x+f(1)$

Now, taking $x\mapsto-x$ and adding will cancel out the $2xf(1)$ term:
$f(-x)\left(1-f(1)\right)=f(x)+(4-2f(1))x+f(1)$

Adding both of these:
$\left(f(x)+f(-x)\right)\left(1-f(1)\right)=\left(f(x)+f(-x)\right)+2f(1)$
$-f(1)\left(f(x)+f(-x)\right)=2f(1)$
This means that either $f(1)=0$ or $f(x)+f(-x)=-2$.

[b]Case 2.1:[/b] $f(x)+f(-x)=-2$

We get $f(-x)=-2-f(x)$. Comparing $P(x,y)$ with $P(-x,-y)$ proves:

\begin{align*}
&\phantom{\implies}f(-x-y)+f(-x)f(-y)=f(x+y)+f(x)f(y) \\
&\implies -2-f(x+y)+4+2f(x)+2f(y)+f(x)f(y)=f(x+y)+f(x)f(y) \\
&\implies f(x+y)=f(x)+f(y)+1\end{align*}

Note that the substitution $g(x)=f(x)-1$ proves that $g$ is additive, but there's no way to prove that it's linear.

Substituting this into $P(x,y)$, we get:
$$f(x)+f(y)+1+f(x)f(y)=f(xy)+2xy+1$$

Now let $h(x)=f(x)+1$. We have:
$$Q(x,y):h(x)h(y)=h(xy)+2xy$$
Since $f(-1)=0$ and $f(x)+f(-x)=-2$, we know that $h(-1)=1$ and $h$ is odd. The latter means that $h(1)=-1$.

Now, $Q(x,1)\implies h(x)=-x$

Therefore, $f(x)=h(x)-1=\boxed{-x-1}$, which indeed works. There are no other solutions in this case.

[b]Case 2.2:[/b] $f(1)=0$

Substituting this into $(3)$, we see that $f(-x-1)=f(x+1)\forall x$, and $x\mapsto x-1$ shows that $f$ is even.
\begin{align*}
P(x,-y)&\implies f(x-y)+f(x)f(-y)=f(-xy)-2xy+1 \\
&\implies f(x-y)+f(x)f(y)=f(xy)-2xy+1
\end{align*}
We know that $f(x+y)+f(x)f(y)=f(xy)+2xy+1$, so we can say that:
$$f(x-y)+2xy=f(x+y)-2xy$$
$$f(x-y)+4xy=f(x+y)$$
Here, taking $x=y$ gives:
$$f(2x)=4x^2-1$$
And doing the transformation $x\mapsto\frac x2$, we get:
$$\boxed{f(x)=x^2-1}$$
which also works.

Thus, our solutions are $\boxed{f(x)=2x-1}$, $\boxed{f(x)=x^2-1}$ and $\boxed{f(x)=-x-1}$. $\square$

Another solution since I re-did this problem without realizing it.

Let $P(x,y)$ be the assertion $f(x+y)+f(x)f(y)=f(xy)+2xy+1$.
$P(x,0)\Rightarrow f(x)+f(x)f(0)=f(0)+1$ so $f(0)=-1$ (else $f(x)=1$, which doesn't work).
$P(1,-1)\Rightarrow f(1)f(-1)=f(-1)\Rightarrow f(-1)=0\vee f(1)=1$

[b]Case 1:[/b] $f(-1)=0$
$P(1,-2)\Rightarrow f(1)f(-2)+3=f(-2)$
$P(-1,-1)\Rightarrow f(-2)=f(1)+3$
So $f(1)^2+2f(1)=0$, thus either $f(1)\in\{0,-2\}$.

[b]Case 1.1:[/b] $f(1)=0$
Let $g(x)=f(x)-x^2+1$, so that $g$ has zeroes at all elements of the set $\{-1,0,1\}$. We get the assertion
$$Q(x,y):g(x+y)+g(x)g(y)-g(x)-g(y)+x^2g(y)+y^2g(x)=g(xy)$$
$Q(x,1)\Rightarrow g(x+1)=g(x)$
$Q(-x,-1)\Rightarrow g(-x-1)=g(x)$
thus $g$ is even, since $g(x+1)=g(-x-1)$.
$Q(x,-y)\Rightarrow g(x-y)+g(x)g(y)-g(x)-g(y)+x^2g(y)+y^2g(x)=g(xy)$ (using evenness)
So $g(x+y)=g(x-y)$, which we obtain when we compare the above with $Q(x,y)$. Setting $x=y$ yields $g(2x)=g(0)=0$, so $g(x)=0$, and then $\boxed{f(x)=x^2-1}$, and it's easy, but a bit annoying to verify that this works.

[b]Case 1.2:[/b] $f(1)=-2$
\begin{align*}
P(x-1,1)&\Rightarrow f(x)-2x+1=3f(x-1)\\
P(-x+1,-1)&\Rightarrow f(-x)=f(x-1)+2x-1\\
&\Rightarrow f(-x)=\frac{f(x)+4x-2}3\\
&\Rightarrow3f(x)=f(-x)-4x-2=\frac{f(x)+4x-2}3-4x-2\\
&\Rightarrow8f(x)=-8x-8\\
&\Rightarrow\boxed{f(x)=-x-1},\\
\end{align*}
which works.

[b]Case 2:[/b] $f(1)=1$
$P(x-1,1)\Rightarrow\boxed{f(x)=2x-1}$, which also works.

Solutions, which all work: $\boxed{f(x)=x^2-1}$, $\boxed{f(x)=-x-1}$, $\boxed{f(x)=2x-1}$

I redid this problem yet again without realizing it.

Let $P(x,y)$ denote this assertion.
If $f(0)\ne-1$, then:
$P(x,0)\Rightarrow f(x)=1$ which doesn't work.
So $f(0)=-1$.
$P(1,-1)\Rightarrow f(1)f(-1)=f(-1)$
If $f(1)=1$ we have:
$P(x-1,1)\Rightarrow\boxed{f(x)=2x-1}$, which is a solution.
Otherwise, assume $f(-1)=0$.
$P(-2,1)\Rightarrow f(-2)f(1)=f(-2)-3$
$P(-1,-1)\Rightarrow f(-2)=f(1)+3$
Substituting $P(-1,-1)$ into $P(-2,1)$, we have $f(1)\in\{-2,0\}$.

[b]Case 1:[/b] $f(1)=0$
$P(x,1)\Rightarrow f(x+1)=f(x)+2x+1$
$P(x,y+1)-P(x,y)\Rightarrow 2x^2+4xy+2y^2+2yf(x)+f(x)=f(xy+x)-f(xy)+2x-1$
Setting $y=\frac1x$ and making $x\ne0$, we obtain $f(x)=x^2-1$ for $x\ne0$. Since $f(0)=-1$, it completes the solution $\boxed{f(x)=x^2-1}$ which fits.

[b]Case 2:[/b] $f(1)=-2$
$P(x,1)\Rightarrow f(x+1)=3f(x)+2x+1$
$P(x,y+1)-3P(x,y)\Rightarrow 2y+3+2yf(x)+f(x)=f(xy+x)-3f(xy)-4xy$
We repeat with $y=\frac1x,x\ne0$, we have $f(x)=-x-1$ for $x\ne0$. Since $f(0)=-1$ this is $\boxed{f(x)=-x-1}$, which is a solution.

Really straightforward problem.

Let $P(x;y)$ denotes the assertion of given functional equation. Note that $P(0;0)$ gives us:
$$ f(0)+f(0)^2 = f(0)+1 \implies f(0)= 1 \quad \text{or} \quad f(0)=-1 $$
Suppose that $f(0)=1$. Then from $P(x;0)$ we have:
$$ f(x)+f(x) =1 +1 \implies f(x)=1 $$
It is easy to see that this function does not work; therefore we can assume that $f(0)=-1$.

Consider $P(1;-1)$:
$$ f(0)+f(1)f(-1)=f(-1)-2+1 \implies f(1)f(-1)=f(-1) \implies f(-1)=0 \quad \text{or} \quad f(1)=1 $$
Suppose that the latter is the case. Then from $P(x;1)$ we have:
$$ f(x+1) +f(x)=f(x)+2x+1 \implies f(x+1)=2x+1 \implies f(x)=2x-1$$
It is easy to see that this function works; thus we can assume that $f(-1)=0$.

We would like to determine the value of $f(1)$. To do so we consider $P(2;-1)$, which gives us:
$$ f(1)=f(-2)-3 $$
On the other hand, $P(1;-2)$ reveals that:
$$ f(1)f(-2)=f(-2)-3 \implies f(1)(f(1)+3)=f(1)+3-3 \implies f(1) =0 \quad \text{or} \quad f(1)=-2 $$
Let's consider the case when $f(1)=0$.

From $P(x;1)$ we have:
$$ f(x+1)=f(x)+2x+1 \implies f(x)=f(x-1)+2x-1 \implies f(x)-2x+1=f(x-1)$$
On the other hand, $P(x,-1)$ gives us:
$$ f(x-1) = f(-x)-2x+1 = f(x)-2x+1 \implies f(x)=f(-x) $$
Now it is quite natural to consider $P(x;-x)$:
$$ -1 +f(x)f(-x)=f(-x^2)-2x^2+1 \implies f(x)^2=f(x^2)-2x^2 +2 \qquad (\star)$$
Also note that $P(x;x)$ gives us:
$$ f(2x) +f(x)^2 = f(x^2) +2x^2 +1 \qquad (\star \star) $$
Subtracting $(\star)$ from $(\star \star )$:
$$ f(2x) = 4x^2 -1 \implies f(x) = x^2 -1 $$
It is easy to check that this function indeed works.

Now comes the painful part, when $f(1)=-2$. Since $f(1)=f(-2)-3$, we must have $f(-2)=1$. Also note that from $P(1;1)$ we have $f(2)=-3$. First of all, observe that from $P(x;1)$ we have:
$$ f(x+1) -2f(x) = f(x)+2x+1 \implies f(x+1) = 3f(x)+2x+1 $$
Now $P(x;2)$ gives us:
\begin{align*}
f(x+2) -3f(x) =f(2x)+4x+1 \\
3f(x+1)+2x+3 =3f(x)+f(2x)+4x+1 \\
9f(x)+6x+3+2x+3 =3f(x)+f(2x)+4x+1 \\
6f(x)+4x+5=f(2x) \implies 6f(-x)-4x+5 = f(-2x)
\end{align*}
It also makes sense now to consider $P(x;-2)$:
\begin{align*}
f(x-2) +f(x)=f(-2x) -4x+1 \\
9f(x-2)+9f(x) =9f(-2x)-36x+9 \\
3(f(x-1)-2(x-2)-1)+9f(x) = 9f(-2x)-36x+9 \\
3f(x-1)-6x+9+9f(x) =9f(-2x)-36x+9 \\
3f(x-1)+9f(x) =9f(-2x)-30x \\
f(x)-2(x-1)-1+9f(x)=9f(-2x)-30x \\
f(x)-2x+1+9f(x)=9f(-2x)-30x \\
10f(x) +1 = 9(-2x)-28 x \\
10f(x)+1 = 9(6f(-x)-4x+5) -28x \\
10f(x) =54f(-x)-64x+44 \\
5f(x) =27f(-x)-32x+22
\end{align*}
Our pain ends here, since now we can relate $f(x)$ and $f(-x)$ using only $x$ and some constants. The final step is to take a look at $P(x;-1)$:
\begin{align*}
f(x-1) =f(-x)-2x+1 \\
27f(x-1)=27f(-x)-54x+27 \\
27f(x-1)=5f(x)+32x-22-54x+27 \\
27f(x-1)=5f(x)-22x+5 \\
9(3f(x-1))  =5f(x)-22x+5 \\
9(f(x)-2(x-1)-1) = 5f(x)-22x+5 \\
9f(x) -18x+9 = 5f(x)-22x+5 \\
4f(x) =-4x-4 \\
f(x) = -x-1
\end{align*}
It is easy to check that this function also satisfies given equations. With all cases being considered, we conclude that only possible functions are $f(x)=x^2-1$; $f(x)=2x-1$ and $f(x)=-x-1$.

Let $P(x,y)$ be the assertion.

[hide=Finding f(0)]$P(0,0) \rightarrow f(0)=1, -1$
Assume $f(0)=1$.
$P(x,0) \rightarrow f(x)=1$
$P(2,2) \rightarrow f(2)=\pm 3$, contradiction.
Thus, $f(0)=-1$.[/hide]

[hide=Cases]$P(1,-1) \rightarrow f(1)=1$ or $f(-1)=0$.
First case, let $f(1)=1$.
$P(x,1) \rightarrow f(x)=2x-1$, which works.
Second case, $f(1) \neq 1, f(-1)=0$.
$P(-2, 1) \rightarrow f(1)f(-2)=f(-2)-3$.
$P(-1,-1) \rightarrow f(-2)=f(1)+3$.
Combining gives $f(1)=0, f(-2)=3$ or $f(1)=-2, f(-2)=1$.
First subcase, let $f(1)=0, f(-2)=3$.
$P(x,-1), P(-x,1)$ gives $f(x)=f(-x)$.
$P(x,-x) \rightarrow f(x)^2+2x^2=f(x^2)+2$.
$P(x,x) \rightarrow f(x)=x^2-1$, which works.
Second subcase, let $f(1)=-2, f(-2)=1$.
$P(x-1,1) \rightarrow f(x)+1=3f(x-1)+2x$.
$P(1-x, -1) \rightarrow f(-x)+1=f(x-1)+2x$.
Then $f(x-1)=-f(-x-1)$ and some manipulations yields $f(x)=-x-1$, which works.[/hide]

[hide=Finish/Summary]Thus, the only solutions are $f(x)=2x-1, x^2-1, -x-1$, which all work. $\blacksquare$[/hide]

Let $P(x,y)$ be the assertation that  $f(x+y)+f(x)f(y)=f(xy)+2xy+1$
 $P(0,0) \Rightarrow f(0)=\pm1$
 If $f(0)=1$ then $P(x,0) \Rightarrow f(x)=1$
 $P(x,y) \Rightarrow 0=2xy \Rightarrow$ contradiction $\Rightarrow f(0)=-1$
 $P(-1,1) \Rightarrow f(0)+f(-1)f(1)=f(-1)-1 \Rightarrow f(-1)(f(1)-1)=0$
[b][u]Case 1[/u]:[/b]  [b]f(1)=1[/b] $\Rightarrow$
$ P(x,1) \Rightarrow f(x+1)+f(x)f(1)=f(x)+2x+1 \Rightarrow f(x+1)=2x+1 \Rightarrow \boxed{f(x)=2x-1}$
[b][u]Case 2[/u]:[/b] [b]f(-1)=0[/b] $\Rightarrow P(-1,-1) \Rightarrow f(-2)=f(1)+3$
$P(-2,1) \Rightarrow f(-2)f(1)=f(-2)-3 \Rightarrow f(-2)=\frac{-3}{f(1)-1} \Rightarrow f(1)+3=\frac{-3}{f(1)-1} \Leftrightarrow $ $f(1)=0$ or $f(1)=-2$
[b][u]Subcase 2.1[/u][/b][b]: f(1)=0[/b] $\Rightarrow P(x,1) \Rightarrow f(x+1)=f(x)+2x+1$
$P(x+1,-1) \Rightarrow f(x)=f(-x-1)-2x-1 \Rightarrow f(x+1)=f(-x-1) \Rightarrow f(x)=f(-x)$ for all $x \in \mathbb{R}$
$P(x,-x) \Rightarrow f(x)^2=f(x^2)-2x^2+2 \Rightarrow P(x,x) \Rightarrow f(2x)+f(x)^2=f(x)^2+2x^2-2+2x^2+1 \Rightarrow \boxed{f(x)=x^2-1}$
[b][u]Subcase 2.2[/u]: f(1)=-2 [/b]$\Rightarrow P(1,1) \Rightarrow f(2)=-3$
$P(x,1) \Rightarrow$ $f(x+1)=3f(x)+2x+1$ [b](1)[/b] $\Rightarrow f(x+2)=9f(x)+8x+6$ [b](2)[/b]
$P(x,-1) \Rightarrow$ $f(x)=f(-x-1)-2x-1 \Rightarrow f(x)=3f(-x)-4x+2$ [b](3)[/b]
Using (1) and (3): $P(x,-x) \Rightarrow$ $f(x)f(-x)=f(-x^2)-2x^2+2 \Rightarrow f(x^2)=\frac{3f(x)^2+4xf(x)+2f(x)-2x^2-4}{3}$[b](4)[/b]
Using (1),(3) and (4): $P(x,x) \Rightarrow f(2x)+f(x)^2=f(x^2)+2x^2+1 \Rightarrow f(2x)=\frac{4xf(x)+2f(x)+4x^2-1}{3}$[b](5)[/b]
Using (2) and (5): $P(x,2) \Rightarrow f(x+2)-3f(x)=f(2x)+4x+1\Rightarrow$ after substituting $\boxed{f(x)=-x-1}$
After checking we see that all 3 solution work.


Prove the inequalities

1)  m_a+m_b+m_c \geq a+b

2)  m_b+m_c  \leq 3/4(a+b+c)

where m_a,m_b,m_c are medians of a triangle of sides a,b,c.

2) Is easy :
Let B' be the midpoint of the side AC, and B'' be the symmetric of B with respect to B'.
Thus BB' = 2m_b < BA + AB'' = c + a.
Similarly 2m_c < b + a.
Adding, we get the result from the triangle inequality.

Pierre.

For the first one:

Let G be mass-center. Then
        GA+GC >= b
        GB + GC>= a
        (GA+GB)/2 - GC >= 0
Sum up these ineqs we have done.

Namdung


show that the set of numbers in the interval [0,1] having decimal expansions using only odd digits is closed. describe this set by a Cantor-set type construction.

Well, if they've got only odd digits, then the only possibility is $\{1\}$, which is of course closed  :lol:

Edit: I'm joking, of course. The closedness is obvious (show that the complement is open).


MATHCOUNTS AMC USA(J)MO USAMO Putnam USAMTS \/closed

You can now hide the forums that you don't usually visit, to reduce some of the clutter on the Forum main page.

The new option is available in the [url=http://www.mathlinks.ro/Forum/usercp.php]options menu[/url]. Once there click on "Hide Forums Panel".

As in the pictures below, for each forum you can:

- Select "View" to have that forum appear on the Forum main page.

- Select "Hide Forum" to have that forum be hidden on the Forum main page.

- Click on the green arrow to copy that forum's setting to all of its subforums.

You must click Submit at the bottom of the page for any changes to take effect.

If you hide a forum but leave visible any of its subforums, then the subforums will still appear on the Forum main page.

You can still visit a forum that you have hidden, by selecting it on the "Jump to:" link on the lower-right of the Forum page.

If you have any questions about this, please post them in this topic.

[quote="tokenadult"]The options pane doesn't display with the same line spacing as the forum list pane when I view the hide forums with Firefox. [/quote]I cannot reproduce this particular problem on my copy of Firefox (version 1.0).  However, there are definitely some issues using this feature in Firefox...in particular, it is [b]very[/b] slow to load.  This is due to the way that Firefox processes Javascript.  However, once the page is completely loaded (and it takes about 2 minutes  :sleeping: on my machine), it does seem to be lined up fine (again, at least on my machine) and the feature [i]does[/i] work.

We'll look into it, but I don't think this is going to be any easy fix (if it's fixable at all).

YAY!!! Thank you! :clap: I now don't see College or Olympiad or non-North American forums, since I've never been there. Let me reiterate: THANK YOU!

No problem on Deer Park Alpha 1, though it takes some time to load; that's why you have multiple tabs loading at one time ;)

:clap: :clap2:

Billy

[quote="tokenadult"]The options pane doesn't display with the same line spacing as the forum list pane when I view the hide forums with Firefox. I'm not sure what forum I'm choosing for that way, so I don't dare make any changes.[/quote]The quickest and simplest solution would be just to use another browser to modify your hide forums settings. Obviously you won't do this often (it beats the whole purpose of it :P) so I recommend you use IE to set up your options.

Firefox has some major issues dealing with Java Script. This is a problem of the browser (on some computers, on mine it loads in about 20 seconds and it works fine), but I don't think there are any problems with either Opera or IE :)

Firefox actually works fine for me...it didn't take long to load at all! :) Thankyou for this new feature, you guys are doing a great job with this forum! :)

[quote="chess64"]Firefox actually works fine for me...it didn't take long to load at all! :)[/quote]Mine work fine too (on both my home computer, my laptop, and my aops computer). However on some computers (especially those not having a high processor speed and a ton of RAM) it might not work. I generally use IE to switch the options, as IE is instant.

[quote="chess64"]Thankyou for this new feature, you guys are doing a great job with this forum! :)[/quote]Thanks! :P We are trying :D

Yep, I use Firefox, but you're right, it is loaded with ram and processing speed. By the way, nice avatar/picture: GO YODA! YODA RULES!

thank you for all the AOPs people who set up this forum, yall r the best! :first:  :clap2:

and go yoda

[quote="Stijn"]Firefox works fine here.

But there's one little bug, in Firefox and IE. The Middle School green arrow copies the setting to all boars, instead of only the Middle School-subforums. The other green arrows work fine.[/quote]What is "boars"? Are you sure you are clicking the middle school arrow and not the whole forum arrow? There's no such bug on my computer(s) ...

[quote="Stijn"]I meant to write 'boards'/'forum'.

Yes, I'm sure clicking the correct arrow. I'm going to 'digg' into the source code a bit, see if I can find the problem.[/quote]Can you post a print screen of the arrow you are clicking and what it does (before and after the click)?

PS How are you planning to "digg into the source code"?!!  :huh:

[quote="Valentin Vornicu"]Can you post a print screen of the arrow you are clicking and what it does (before and after the click)? [/quote]

It doesn't work for me either. See attached.

[quote="Stijn"]For some reason, he thinks of the MATHCOUNTS forum having both id 132 (correct) as 301 (incorrect/nonexistant).
The Middle School green arrow does javascript:branch_set('styles',298,301), but since there is no board 301, the keeps changing the View/Hide option to the bottom.
The same with the MATHCOUNTS green arrows, which does javascript:branch_set('styles',132,301).[/quote]Can you post a print screen?!! Thanks. I have not experienced this problem on any computer(s) ...

[url=http://download.stijnshome.be/before.png]Before[/url] and [url=http://download.stijnshome.be/after.png]after[/url] I clicked on the Middle School green arrow. Notice everything below changes to 'Hide Forum'. Notice the status bar javascript-message too.

I am having no such problems on none of my installed browsers ... this is very weird. What browsers are you guys using?

Internet Explorer. He's using Firefox. See another screenshot attached.

[quote="Stijn"]I have the problem on both Firefox 1.0.4 and IE6. So I doubt this is an browser-related issue...[/quote]Well I have those 2. And I have tried reproducing the bug on 3 different computers, and with 2 different accounts (my account, and a regular user account, just to be on the safe side) and I could not do it ... it might be a problem with your java installation. Do you have the latest java rte?

I didn't notice the arrow thingy and changed each and every one of the forums I didn't want to see manually (that's all of the forums and subforums in the olympiad and college math sections). Bleh. Oh well...thanks so much for this feature. The forums loading and then collapsing was beginning to drive me crazy.

Valentin, if you don't experience this, YOUR java is not working ;)

Where the source code writes

javascript:branch_set('styles', 298, 301);

it should be

javascript:branch_set('styles', 298, xxx);

with xxx the last forum in that category, I don't know which forums admins have there but for normal users it is 132.

[quote="Valentin Vornicu"]Anyway, besides this (minor) bug, are there any important bugs you guys have noticed? In the end just avoid hiding the middle school forums, or do it manually (there are only like 5 subforums :P) ...[/quote]

I noticed another strange thing, though it may be because of my stupidity.

In high school, I hided only:
Getting Started

And the forums that I see there now in high school are:
Grading Issues
Other US Contests & Programs
International Contests & Programs

Which do not belong there at all. This also prevents me from accessing the forum I am moderating :o which cannot be intended?

[edit] when I then click on the category, I get the normal selection. It would be nice if I could get it main screen though ;) [/edit]

Valentin, could you please make that posts from hidden forums do not appear at "View posts since last visit" link?

[quote="Peter VDD"]I noticed another strange thing, though it may be because of my stupidity.

In high school, I hided only:
Getting Started

And the forums that I see there now in high school are:
Grading Issues
Other US Contests & Programs
International Contests & Programs

Which do not belong there at all. This also prevents me from accessing the forum I am moderating :o which cannot be intended?

[edit] when I then click on the category, I get the normal selection. It would be nice if I could get it main screen though ;) [/edit][/quote]If you won't post a print screen it is very hard to understand what is the problem you are experiencing ....

You have probably hidden only the mother forums, not all the subforums too ... :)

@Bojan: no, that will not happen very soon. Partially because the hidden forums thing, is just what it says it is: hiding the forums, so that your index.php pages are not as crowded. But I want everyone to still be able to click on links to the hidden forums or jump into them using the jumpbar.

[quote="Valentin Vornicu"]You have probably hidden only the mother forums, not all the subforums too ... :)[/quote]I'm not THAT stupid ;)

Here are the print screens, the error in the index and the correct display in the category-view.

I discovered why it is. There are a couple of forums (grading issues, other US contests, classes information - very annoying forum :?) that don't have buttons, and therefor can't be hidden. (and I still had to roll them out - this is quite confusing)

Could you get those without a button fixed soon? :)

Thanks for this great system by the way, it is indeed this we wanted when the roll-out system came. :)

(besides, is it normal that the configuration page takes >1 minute to display? :?)

[quote="Bojan Basic"][quote="Valentin Vornicu"]
@Bojan: no, that will not happen very soon. Partially because the hidden forums thing, is just what it says it is: hiding the forums, so that your index.php pages are not as crowded. But I want everyone to still be able to click on links to the hidden forums or jump into them using the jumpbar.[/quote]
When I come to the forum I click on ""View posts since last visit" and read new posts. I think it is logical that if one hid some forums then doesn't want to read new posts from them, but if you have different opinion - no problem, it's your choice. Thank you anyway.[/quote]I said that it will not happen very soon. You guys should understand that it´s not very easy to add all these features and a lot of effort is put into each of them (besides all the other stuff we have to do like jobs, college, classes and stuff :) ).

Peter: some of the forums cannot and will not be hidden on purpose. We want people to be able to see those forums all the time (I am saying that it´s not a bug).

[quote="Peter VDD"]:o

why do non-US residents need to see "other US contests"?? :huh: we can't participate in any of them?

Same for "grading issues", what's the use of it? The classes I can understand, since it is a source of income to you, but those other 2 forums seem completely useless to me, especially since they focus on US-things in which I cannot participate.

It is a sad evolution, and excuse me if I'm wrong, but I get more and more the impression that the site gets mainly designed for americans... and if you're not, too bad. :|[/quote]I think that you are over reacting :dry: . As I told you, most of you [u]have absolutely no idea how much amount of work[/u] is spent such that you could have all this. Of course nobody is ever fully satisfied, but we are doing out best  :?

Up until a couple of days ago you coudn't hide any forums. Now you can hide 95% of them, but I am making modifications for US residents only ....  :| Jeez ...

Has it [b]ever[/b] crossed your mind that some of the forums had simply been forgotten on the setting "not able to hide"? Of course not, it's easier to blame it on the Romanians that do work only for the US-residents :no:

PS Just for the record you need not be a US-resident to participate in many of the US competitions. I know a bunch of  Romanians that took AMCs, and there are 3 Canadian winners of the USAMO this year on the forum.

PPS If you can't participate in a competition that doesn't mean that you can't or shouldn't check that competitions' forum ... or you shouldn't be ever looking in the Putnam forum, right?

my screen just doesn't let me have the "show" or "not show" buttons... do you know why?? I'm using Internet Explorer, and the buttons showed the first time but now they don't show. is there a problem with my computer? Thanks. :)

sorry if that image is too big but I wasn't sure how to make it so you could download the pic. but as you can see it doesn't work :(

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=13[/img]

Lol... Valentin, I think you misunderstood me a bit ;)

I'm very thankful for the modification made here, as I've posted some posts above! I was just replying to what you said, that it is _intended_ that those forums cannot be hidden.

First you say:
[quote="Valentin Vornicu"]Peter: some of the forums cannot and will not be hidden on purpose. We want people to be able to see those forums all the time. (I am saying that it´s not a bug).[/quote]
Then you say:
[quote="Valentin Vornicu"]Has it [b]ever[/b] crossed your mind that some of the forums had simply been forgotten on the setting "not able to hide"?[/quote]
-> I don't get which of those it is now :)

But again, don't get me wrong. I don't "blame" you for anything! I just wonder why "purely american" forums should be on the list "cannot be hidden by anyone"...

...And eum, perhaps that reply is a bit overreacting as well ;) I have a lot of respect and thanks to your work here. Sorry if my previous message wasn't clear in that. :oops:

Ive also got a similar problem:

http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=16

Ok, as soon as I am back from the IMO I will take a look to check if all the forums that are supposed to be able to be hidden are actually. For now, I have to much things to do here in Mexico :D :juggle:  :play_ball:

[quote="Peter VDD"]Same for "grading issues", what's the use of it?[/quote]

I assume it's just a temporary oversight, as it clearly is useful only to the USAMTS participants here.

I think this is an excellent feature, I don't get what you guys are talking about with all that javascript error fancy things (I know, awful grammar  :P ).

Now, I barely have to scroll to get to the forums I want.  :D

And you can't hide the other US contests and programs?  :huh:

May you explain your rationale behind prohibiting the Calculus/Olympiad/College links from being hidden? Not be to disrespectful; I appreciate your work, and I'm just curious as to why they are forced when there are easier ways to get to the desired forums.

I don't believe they are prohibited form being hidden. I have them hidden on my account. Make sure you hide them them in Profile->Hide Forums.


A deli sandwich consists of one type of bread, one type of meat, and one type of sauce. The deli offers wheat, rye, and white bread; ham, turkey, roast beef, and tuna; ranch and southwest chipotle sauce. Javier is allergic to rye bread, tuna, and southwest chipotle sauce. He tells the server to assemble a random deli sandwich. What is the probability that Javier will suffer an allergic reaction?

There are 3 types of bread, 4 types of meat, and 2 types of sauce. He is allergic to one kind of bread, one kind of meat, and one kind of sauce. The probability that he doesn't have an allergic reation is thus $ \frac {2}{3}\cdot{\frac {3}{4}}\cdot{\frac {1}{2}} \equal{} \frac {1}{4}$. Therefore, the probability that he does suffer an allergic reation is $ 1 \minus{} \frac {1}{4} \equal{} \boxed{\frac {3}{4}}$

To people like me with food allergies, this problem seems partially disrespectful.

[hide=Solution]There are $3\cdot2\cdot1=6$ ways for him to choose a sandwich without any of his allergens, and $4\cdot3\cdot2=24$ ways to make a sandwich. So, the probability of him suffering an allergic reaction is $1-\frac{3\cdot2\cdot1}{4\cdot3\cdot2}=\boxed{\frac{3}{4}}$.[/hide]

I hope he's carrying some epinephrine or something else like that.

why why why did javier tell the server to make a random deli sandwich

[hide=hhhhhhh]so there are 3*4*2 different deli sandwiches
2*3*1 deli sandwiches that wont trigger an allergic reaction
6/24=1/4
subtract from 1
then there's 3/4 chance that he suffers an allergic reaction[/hide]

what was bro thinking, bros literally going to die
[hide=smh]
3*4*2 = 24 total sandwiches
2*3*1 = 6 where he lives
18 sandwiches where he dies[/hide]


[quote="towersfreak2006"]If $X$ and $Y$ are metric spaces and $f\;: \;X\;\rightarrow\;Y$ is a mapping between them, show that the following statements are equivalent:
a) $f^{-1}(B)$ is open in $X$ whenever $B$ is open in $Y$.
b) $f^{-1}(B)$ is closed in $X$ whenever $B$ is closed in $Y$.
c) $f(\overline{A})\subseteq\overline{f(A)}$ for every subset $A$ of $X$.[/quote]

It has been done in one of my very last posts.


What is the largest 2 digit prime factor of $ {200\choose 100}$?


geometry
circumcircle
geometric transformation
reflection
geometry unsolved

We are given a square $ ABCD$. Let $ P$ be a point not equal to a corner of the square or to its center $ M$. For any such $ P$, we let $ E$ denote the common point of the lines $ PD$ and $ AC$, if such a point exists. Furthermore, we let $ F$ denote the common point of the lines $ PC$ and $ BD$, if such a point exists. All such points $ P$, for which $ E$ and $ F$ exist are called acceptable points. Determine the set of all acceptable points, for which the line $ EF$ is parallel to $ AD$.

First, we show that P must lie on the circumcircle of $ DOC$, where O is the center of the square.

Assume $ EF$ is parallel to $ AD$. Then $ EF \perp CD$ and $ CE \perp DF$. Therefore, $ E$ is the orthocenter of $ FDC$, and thus $ DE \perp CF$. Therefore, $ \angle DPC\equal{}90$, and this along with $ \angle DOC\equal{}90$ yield the desired result.

Now, the acceptable points on this circle are merely those points that are [b]not $ O,C,D$[/b] or the reflection of $ O$ about $ CD$. We will show that all of the points on the circle that are not $ O,C,D$ or the reflection of $ O$ about $ CD$ work.

$ CO \perp DF$ and $ DP \perp CF$ together imply that the intersection of $ CO$ and $ DP$, $ E$, is the orthocenter of triangle $ DFC$. Therefore, $ EF\perp CD$ and we are done.

it suffices to prove $\angle CPD=90$
besides,I think that the answer can include O.


geometry
geometric transformation
rotation
analytic geometry

The disk of m mass has a rotation axis. Find the lagrangian of the system.
The generalized coordinate is the angle $\phi$
[url=http://img53.imageshack.us/my.php?image=discku6.jpg][img]http://img53.imageshack.us/img53/4086/discku6.th.jpg[/img][/url]


$ ABCD$ is a quadrilateral such that $ AB \equal{} AD$. If $ \angle ACB \equal{} \angle ACD$, prove that $ ABCD$ is cycle quadrilateral.

Sorry for my bad English  :D .

Are you sure this is true?
 [geogebra]1836c2422573c312791e2e5bffdabefd8015762f[/geogebra]


In my Latex document for an AoPS class final, I used eqnarray with the equation numbers for one of the problems.  If I use eqnarray again for a later problem, how can I get the equation numbers to start back at (1)?  Or should I just have everything continuing throughout the document, even though the problems are unrelated?

Use the command \setcounter{equation}{0} before begining the new eqnarray enivronment.


the chickens and pigs in Farmer McCoy's barn have a total of 50 heads and 170 legs. How many pigs are in the barn?

Let $ x\equal{}\text{chickens}$, $ y\equal{}\text{pigs}$.
$ \left\{\begin{array}{l}
x\plus{}y\equal{}50 \\
2x\plus{}4y\equal{}170
\end{array}\right.$
$ \iff (x, y)\equal{}(15, 35)$

Thus, $ \boxed{35}$.


Diophantine equation
number theory unsolved
number theory

Given the Pell equation $ x^2 \minus{} dy^2 \equal{} 1$ where $ d$ is not a perfect square and a fixed positive integer $ k$, prove that you can find infinitely many solutions with $ y$ divisible by $ k$.

put $ y\equal{}y'k$ then the pell equation will be:

$ x^2\minus{}(dk^2)y'^2\equal{}1$

put $ d'\equal{}dk^2$ then $ d'$ isn't a perfect square and so the pell equation $ x^2\minus{}d'y'^2\equal{}1$ has infinitely many solutions.


Putnam
function
induction
combinatorics unsolved
combinatorics

Let $ \mathbb{P}_n$ be the set of subsets of $ \{1,2, \cdots, n\}$. Let $ c(n,m)$ be the number of functions $ f : \mathbb{P}_n \to \{1,2, \cdots, m\}$ such that $ f(A \cap B) \equal{} \mbox{min}\{f(A), f(B)\}$. Prove that $ c(n,m) \equal{} \sum_{j \equal{} 1}^m j^n.$

I don't really get the notation used in this question, what does $ f(A \cap B) \equal{} \mbox{min}\{f(A), f(B)\}$ mean? And how do you approach this question?

Thank you!

The minimum of two numbers is another number. That makes this a meaningful equation, and we want to find all functions on sets satisfying it.

The nicest way to prove a counting formula is to construct a bijection. For a typical term $ j^n$, what does $ j^n$ count? Well, the obvious choice is functions from $ \{1,2,\cdots,n\}$ (or some other set of size $ n$) to $ \{1,2,\cdots,j\}$ (or some other set of size $ j$).

OK, what if we started with a function $ g$ from $ \{1,2,\cdots,n\}$ to $ \{1,2,\cdots,j\}$; we want to turn that into an $ f$ satisfying our property. Clearly*, we should have $ f(\{k\})\equal{}g(k)$. Then $ f(\{1,2\})\equal{}\dots$ Hmm, maybe this isn't so obvious- but if we look instead for $ f(A\cup B)\equal{}\min\{f(A),f(B)\}$, we can build everything easily, and $ f(A)\equal{}\min_{x\in A}g(x)$ for nonempty $ A$. How to go from unions to intersections? DeMorgan's laws; just flip to the complement $ f(\{1,2,\cdots,n\}\setminus k)\equal{} g(k)$.
* OK, that was really just a guess- but one-element sets are usually a good place to start when building a set function.

That's all very nice, but we've got two problems: the functions we get for $ j$ are a subset of those for $ j\plus{}1$ rather than being disjoint, and we haven't accounted for the value of $ f$ on the whole set $ S$. How to deal with that? The obvious choice is to just pick $ f(S)$ and see what that does to the rest of the problem. Since $ f(A)\equal{}f(S\cap A)\equal{}\min\{f(S),f(A)\}$ for any $ A$, $ f(S)$ is at least as large as any other $ f(A)$. If $ f(S)\equal{}j$, all other values are in $ \{1,2,\cdots,j\}$; pick $ n$ of them to be $ f(S\setminus k)$, and we have $ j^n$ functions with $ f(S)\equal{}j$. Sum over $ j$, and we get the desired formula $ c(n,m)\equal{}\sum_{j\equal{}1}^m j^n$.

That whole post was pretty much stream-of-consciousness; it's the process I just went through to solve this.

Thanks for that but what about using induction to prove it?

This is what I have so far:

Say you had a function $ f : \{1,2,3\} \to \{1,2,3\}$ then there exist $ 3^3$ different functions.

If you had $ f: \{1,2...n\} \to \{1,2...m\}$ then there exist $ m^n$ different functions.

If you had $ f: \mathbb{P}_n \to \{1,2...m\}$ then there exist $ m^{2^n}$ different functions.

So out of these $ m^{2^n}$ functions we need to find how many satisfy the restriction $ f(A \cap B) \equal{} \mbox{min}(f(A) , f(B))$

So then using induction:

Base Case:

Let $ m \equal{} 1$

$ f: \mathbb{P}_n \to \{1\}$

$ c(n,1) \equal{} 1 and 1^{2^n} \equal{} 1$

Which is true since the only possible function satisfies the restriction $ f(A \cap B) \equal{} \mbox{min}(f(A) , f(B))$

Inductive Hypothesis:

Assume it is true for $ m \equal{} k$

So $ c(n,k) \equal{} \sum_{j\equal{}1}^k j^n \equal{} 1\plus{} 2^n\plus{}3^n \plus{} ... \plus{} k^n$ is true

Proof:

Need to prove it is true for $ m \equal{} k\plus{}1$

$ c(n,k\plus{}1) \equal{} \sum_{j\equal{}1}^{k\plus{}1} j^n \equal{} 1\plus{}2^n\plus{}3^n\plus{}...\plus{}k^n\plus{}(k\plus{}1)^n$

$ \therefore c(n,k\plus{}1) \equal{} c(n,k) \plus{} (k\plus{}1)^n$

Then what?

Then you ask which functions are the new ones (they're the ones with the value $ k\plus{}1$ somewhere), which puts you in pretty much the same place as my solution, only possibly handled in a different order. I don't see a way to do this without that explicit bijection for each $ j^n$.


a) Find all polynomials of the form
        $ f(x) = x^5 - 3x^4 + 2x^3 + ax^2 + bx + c $
if it is known that $f(x)$ is divided by $(x-1)(x+1)(x-2)$.
b) Let $P(x), Q(x), R(x)$ be plolynomials with real coefficients of degree 3, 2, 3 respectively satisfying condition $(P(x))^2 + Q(x)^2 = R(x)^2 $. Find the minimal possible number of real roots (counted with multiplicity) of polynomial $T(x) = P(x)Q(x)R(x)$.

[b]a[/b]: Easy to find out the result
[b]b[/b]: At least 6 ;)


Perhaps it is my ignorance of perhaps it is rumor but...

I have heard that you can qual for USAMO (MOP etc.) as a first yr. undergrad. Is this rite? I was wondering why it said grade 13 on some of those USAMOers on the list...

Sorry, but you can not take the USAMO in college.  That is because, as you know, you must take the AMC 10 or AMC 12 to qualify.  The AMC 12 is only open to people in 12th grade, or under.  Also, I checked and there are no people in 13th grade who qualfied to take the USAMO.

Silly silly me.. I should have checked earlier.. Blasted rumors... Sorry for wastng your time..

Peace,
IW.OAV

Hah! I was right! I think... Look at Eunice Kim here

http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2001-ua/01usamoqualstate.html

From the 2001 olympiad... Eunice is a pretty tite name...

On a completely random aside, I enjoy cheese and my focus is brokus....

Peace,
IW.OAV  :twisted:

Hm...That's peculiar.  Perhaps there are some exceptions.  Also, I have heard people who take the year off after finishing high school, so maybe that's what "13" is referring to.

-interesting_move

Yeah take the year off just to do USAMO... very chic indeed.

IW.OAV

No, that might not be the reason for taking a year off.  It could be that you want to take a break from all the studying in high school. Or, you might just do the USAMO...lol

-interesting_move

While extremely bored, I was checking out the Exeter Academy website.  Apparantly kids go to Exteter after they finished high school for one year.  That can also be another thing.

some kids in india go for a 13th grade. it doesnt mean theyre held back or anything. i dunno why though. hmmmmmmm. i never did read that arml math jam. i should do that

The 13th graders taking the USAMO are typically Canadians who choose to take that additional year of precollege schooling.  At least, that's what I always heard.

There's also the possibility of a typo/misbubbling/scanning error.  The AMC is hardly perfect in these things.

[quote="Lord Venom"]JONGMIN BAEK 9 CUPERTINO HS CUPERTINO CA
AARON KLEINMAN 11 CUPERTINO HS CUPERTINO CA

ch-ching[/quote]

And how exactly does this have anything to do with the present topic?  Not being mean or anything, just confused. :?

Those are the people from his school who qualified for the USAMO 2001, but I don't understand it either.

Ooh, I think Simon just posted the shortest message yet (see "languages" under Etc. if you don't understand, but since Simon Reads All, he probably does know...)

I was thinking about that when I read your other post. It may have been the shortest post, except I believe Chinaboy posted LOL with no period, so that would be tied in terms of the number of characters.


analytic geometry
geometry
trapezoid
algorithm
area of a triangle

Let points A and B have coordinates (0,1) and (4,0) respectively.  There is a point C in the first quadrant that lies on the line y = x + 1 and such that the area of triangle ABC is 20 square units.  Find the coordinates of C.
hint:
[hide] use a determinant for the area of a triangle...[/hide]

[quote="Apprentice"]hint:
[hide] use a determinant for the area of a triangle...[/hide][/quote]
There's a way to solve this problem even if you don't know determinants.
[hide="Hint"]
$A (0,1)$ is on the line $y=x+1$.
So if we regard $AC$ as the base of the triangle, the height is the distance between $B(4,0)$ and the line $y=x+1$.
[/hide]

[quote="Apprentice"]
 use a determinant for the area of a triangle...[/quote]
How can you expres the area of a triangle $ABC$ as a determinant with $A(x_{A},y_{A}),B(x_{B},y_{B}),C(x_{C},y_{C})$. Is there a way to obtain it?

See my PM. It's kind of long and is in Romanian, so I won't put it here. There are two ways to prove it:

1. the formula for distance (from $A$ to $BC$, which is an altitude in our case);

2. consider the trapezium formed by $AB$ and $Ox$ (with a right angle) and so on... sum up the areas of the three trapeziums. (or smth like this)

OK. I'm just going to post the solution now.
[hide="Answer"] $\boxed{C(8,9)}$
I think. :D [/hide]
[hide="Solution"] Call $C(x,x+1)$. $AC$ is the base of length $x\sqrt{2}$. The height of the triangle is the distance from $B$ to the line $y=x+1$, which is $\frac{5\sqrt{2}}{2}$. So the area of the triangle is $\frac{1}{2}\times\frac{5\sqrt{2}}{2}\times{x\sqrt{2}}$.
$\frac{5x}{2}=20 \Longrightarrow x=8$. So the coordinates of C are $\boxed{\boxed{(8,9)}}$ :D [/hide]

[hide]Use Shoesting Algorithm:
A=(0,1), B=(4,0), C=(x,x+1)

0     1
4     0
x     x+1
0     1

First sum is 4(x+1)+x=5x+4
Second sum is 4

Since x is positive, 5x+4 is positive
$\frac{(5x+4)-4}{2}=\frac{5x}{2}=20$
$x=8$
$y=9$
$\boxed{(8,9)}$[/hide]

[quote="mathnerd314"]Could we use like Shoelace or something?[/quote]
I used Shoelace.

[quote="math92"]i took mathnerd314's idea, kind of.
[hide]
what i did was graph the two points, and the line.  I took random points on the line and found the area of it using the shoelace method. this is not a very good method, theres probably a better way to do it.

answer: (8,9)[/hide][/quote]
Guess and check with shoestring? Variables are definately better with larger numbers.


limit
function
real analysis
real analysis solved

$F(x)$ can be rewritten as $(3x^5-20x^3+60x)/32$. Any fixed point for iteration of this function must be a root of $F(x)=x$, or $3x^5-20x^3+28x=0$. This has five roots: $0, \pm\sqrt{2}, \pm\sqrt{\frac{14}3}$. Of these, $0$ and $\pm\sqrt{\frac{14}3}$ are unstable and $\pm\sqrt{2}$ are stable. (This can be seen by drawing the graph of $y=F(x)$ and seeing where and how it crosses the graph of $y=x$.)

$F'(x) = \frac{15}{32}(x^2-2)^2 \ge 0$. From this it follows that the sequence cannot jump over any of the five roots referred to above.

If $x_0 < -\sqrt{\frac{14}3}$, then $x_n \to -\infty$.
If $x_0 > \sqrt{\frac{14}3}$, then $x_n \to \infty$.
If $-\sqrt{\frac{14}3}<x_0<-\sqrt{2}$, then $x_n \to -\sqrt{2}$ from below.
If $-\sqrt{2}<x_0<0$, then $x_n \to -\sqrt{2}$ from above.
If $0<x_0<\sqrt{2}$, then $x_n \to \sqrt{2}$ from below.
If $\sqrt{2}<x_0<\sqrt{\frac{14}3}$, then $x_n \to \sqrt{2}$ from above.

With $x_0=1.5$, we are in the last situation, and $x_n \to \sqrt{2}$ from above.

However, $F'(\sqrt{2}) = 0$ and $F''(\sqrt{2}) = 0$. Thus, the convergence is third order. Let $r_n = x_n-\sqrt{2}$. For small enough $r_n$, we have that $r_{n+1}$ is approximately $\frac54 (r_n)^3$. The sequence converges with extreme rapidity. I haven't nailed down all the details of the following, but it appears true that $r_n<10^{-3^n}$ for all $n$, so that $r_{2004} < 10^{-3^{2004}}$, which is quite an absurd number.

Ok, this is the answer ! The convergence order is of order 3 as you have proved.
Regards,Proposer


algebra
polynomial
inequalities
inequalities unsolved

Let p(x) be a polynomial with positive coefficients. Prove that if $ P(1/x) \geq 1/(p(x))$ holds for x=1, then it holds for all x>0

Plugging $ x\equal{}1$ into $ P(\frac{1}{x}) \geq \frac{1}{P(x)}$ we see that $ P(1)^2 \geq 1$
But by Cauchy-Schwarz we have that for all $ x>0$ we have $ P(x)P(\frac{1}{x}) \geq P(1)^2$   :lol:


Find $ m\in \mathbb{R}$ such that the equation $ z^3\plus{}i(z^2\minus{}1)\minus{}mz\equal{}0$ has real roots .

[hide="Not sure.."]$ z \equal{} a \plus{} bi$

$ a^3 \plus{} 3a^2bi \minus{} 3ab^2 \minus{} ib^3 \plus{} i(a^2 \plus{} 2abi \minus{} b^2 \minus{} 1) \minus{} ma \minus{} mbi \equal{} 0$

$ (a^3 \minus{} 3ab^2 \minus{} 2ab \minus{} ma) \plus{} i(3a^2b \plus{} a^2 \minus{} b^3 \minus{} b^2 \minus{} 1 \minus{} mb) \equal{} 0$

$ 3a^2b \plus{} a^2 \minus{} b^3 \minus{} b^2 \minus{} 1 \minus{} mb \equal{} 0$

Since $ z$ is real, $ b \equal{} 0$.

$ a^2 \equal{} 1$

$ a \equal{} \pm 1$

$ a^3 \minus{} 3ab^2 \minus{} 2ab \minus{} ma \equal{} 0$

Since $ b \equal{} 0$ and $ a \equal{} \pm 1$ we get $ \boxed{m \equal{} 0}$.[/hide]

[hide=" Then I got a different solution"]We want $ z$ to be real. So we must have that $ z^2 \minus{} 1 \equal{} 0$ so $ z \equal{} \pm 1$. Plugging back in we have $ m \equal{} \pm 1$.[/hide]

All three roots can never be all real, since their product is $ i$. If we only want at least one root real, be that $ x$, then $ x(x^2 \minus{} m) \equal{} i(1 \minus{} x^2)$ requires $ x \equal{} \pm 1$ and so $ m \equal{} 1$.


One leg of a right triangle has a length of 9 m. The other sides have lengths that are consecutive integers. Find the number of meters in the perimeter.

[hide="Solution if you didn't memorize Pythagorean Triples"]Let us call the remaining to sides $ x,y$ with $ x$ being the length of the leg and $ y$ being the length of the hypotenuse. Then,
\[ y^2 \minus{} x^2 \equal{} 9^2
\]
. From this, we get
\[ (y \minus{} x)(y \plus{} x) \equal{} 9\cdot9
\]
But we know $ y \equal{} x \plus{} 1$ as the integers are consecutive and the hypotenuse, $ y$, must be the largest.
\[ (1)(2x \plus{} 1) \equal{} 81
\]
But multiplication by $ 1$ doesn't matter.
\[ 2x \plus{} 1 \equal{} 81
\]
Solving the equation we get $ x \equal{} 40$ and $ y \equal{} 41$. Hence, the perimeter is $ 9 \plus{} 40 \plus{} 41 \equal{} 90$.[/hide]

[hide="Solution if you did memorize the Pythagorean Triples"]We know the measures of the triangle are $ (9,40,41)$. Thus, the perimeter is $ 9 \plus{} 40 \plus{} 41 \equal{} 90$.[/hide]

Anyway, I would highly recommend memorizing Pythagorean Triples upto a certain degree which you choose.[/hide]


Prove that

$\sum_{x=1}^{\infty}\left(\frac{1}{n}\right)^x=\frac{1}{n-1}$

Where $n>1$ and is an integer

I don't think I'm allowed to post Sigma problems in Basics or MOEMS, so this'll have to suffice.

[hide]The first term is $1/n$, the ratio is $1/n$, so the sum is $\frac{1/n}{1-1/n}=\frac{1/n}{(n-1)/n}=\frac{1}{n-1} \ \ \ \ \blacksquare$[/hide]

You didn't prove anything, you just used a formula. You're supposed to prove the formula.

As for that, I don't know how to do it. It seems intuitive, though; say $n=2$. You have the first $\frac{1}{2}$, then the sum of the rest will equal $\frac{1}{2}$. I can't prove that, but it's just intuitive, and you can extend it out far enough pretty quickly that you see that it must approach $\frac{1}{2}$.

I thought there weren't supposed to be any infinate series in middle school sections, but I guess these are okay.

yea, but it wasn't a very complicated series either.

[quote="nat mc"]$s=\left(\frac1n\right)^1+\left(\frac1n\right)^2+\left(\frac1n\right)^3+\left(\frac1n\right)^4+\cdots$

$ns=1+\left(\frac1n\right)^1+\left(\frac1n\right)^2+\left(\frac1n\right)^3+\left(\frac1n\right)^4+\cdots$

$(n-1)s=1$

$s=\frac1{n-1}$[/quote]

Maybe it's late at night and my brain's not working properly, but

How did you get from the 2nd step to the 3rd step?

Just subtract the first equation from the second, and all cancel except the 1

Is that the formula for all geometric series like this?

I didn't realize the equation subtracting from the second to the third step either, must be just too early

Right, that makes sense. It's like getting a fraction from repeating decimals.

yeah, thats pretty much the exact way you do repeating decimals, and it proves the sum for all infinite series


calculus
integration
real analysis
real analysis unsolved

Try develop $\frac{1}{1-e^x} = \frac{e^x}{1-e^{-x}}= e^x \sum_{n=0}^{\infty}e^{-nx}$ and then integrate each term separately by redeveloping $e^x$


Let $a,b,c$ be positive reals s.t. $a^2+b^2+c^2=1$. Prove the following inequality \[ \sum \dfrac{a}{a^3+bc} \geq\dfrac{23}{7}. \]

That's false : take $a=b$ close to $0$ and $c$ close to $1$
:cool:


Yeah two people is what we discussed, and it's better anyway. Btw, I accidentally hit 3 in the poll, so consider that.

here is a better poll:

HOW MANY TIMES HAS THIS QUESTION BEEN POLLED??!?!?!??!!?

I actually think that two is better, but then we might have a 16 team competition.

only one smart person is needed. if you have too many people u have a wide mem expansion