do you know anybody who solved this problem?

binaj wrote: do you know anybody who solved this problem? I'm sure someone solved it, but I didn't and neither did the rest of the Austrian team...

Martin N. wrote: Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that for all $x\mbox{, }y\in\mathbb{R}$, we have \[f(x+y)+f(x)f(y)=f(xy)+(y+1)f(x)+(x+1)f(y)\mbox{.}\] (4th Middle European Mathematical Olympiad, Individual Competition, Problem 1) Let $P(x,y)$ be the assertion $f(x+y)+f(x)f(y)=f(xy)+(y+1)f(x)+(x+1)f(y)$ $P(0,0)$ $\implies$ $f(0)^2=2f(0)$ and so $f(0)=0$ or $f(0)=2$ If $f(0)=2$, $P(x,0)$ $\implies$ $f(x)=x+2$ which is not a solution. So $f(0)=0$ $P(1,-1)$ $\implies$ $f(1)f(-1)=3f(-1)$ and so $f(-1)=0$ or $f(1)=3$ If $f(1)=3$, $P(x-1,1)$ $\implies$ $\boxed{f(x)=3x}$ which indeed is a solution. If $f(1)\ne 3$, then $f(-1)=0$ $P(-x,1)$ $\implies$ $f(-x+1)=(3-f(1))f(-x)+(-x+1)f(1)$ $P(x,-1)$ $\implies$ $f(x-1)=f(-x)$ $P(x-1,-1)$ $\implies$ $f(x-2)=f(-x+1)$ And so $f(x-2)=(3-f(1))f(x-1)+(-x+1)f(1)$ But $P(x-2,1)$ $\implies$ $f(x-1)=(3-f(1))f(x-2)+(x-1)f(1)$ Adding these two lines gives $(2-f(1))(f(x-2)+f(x-1))=0$ If $f(1)\ne 2$, this implies $f(x-2)=-f(x-1)$ and so $f(x+1)=-f(x)$ and $f(1)=0$ But then $P(x,1)$ $\implies$ $f(x+1)=3f(x)$ and so $-f(x)=3f(x)$ and $\boxed{f(x)=0}$ $\forall x$ which indeed is a solution. If $f(1)=2$, then $P(x,1)$ $\implies$ $f(x+1)=f(x)+2x+2$ and so : $f(x+n)=f(x)+2nx+n^2+n$ So $f(n)=n^2+n$ Then $P(\frac pq,q)$ $\implies$ $f(\frac pq)=(\frac pq)^2+\frac pq$ and we got $f(x)=x^2+x$ $\forall x\in\mathbb Q$ Let then $g(x)=f(x)-x^2-x$ such that $g(x)=0$ $\forall x\in\mathbb Q$. The original equation becomes : New assertion $Q(x,y)$ : $g(x+y)+g(x)g(y)+g(x)(y^2-1)+g(y)(x^2-1))=g(xy)$ $Q(x,1)$ $\implies$ $g(x+1)=g(x)$ Using this property and subtracting $Q(x,y)$ from $Q(x,y+1)$, we get $g(x)(2y+1)=g(xy+x)-g(xy)$ So $g(x+y)=g(x)+g(y)(\frac{2x}y+1)$ $\forall x,y\ne 0$ So $g(x)+g(y)(\frac{2x}y+1)=g(y)+g(x)(\frac{2y}x+1)$ and so $\frac{g(y)}{y^2}=\frac{g(x)}{x^2}$ $\forall x,y\ne 0$ So $g(x)=ax^2$ and so $g(x)=0$ and $\boxed{f(x)=x^2+x}$ which indeed is a solution. Hence the three solutions of the equation : $f(x)=0$ $f(x)=3x$ $f(x)=x^2+x$

A bit different solution $ f(x+y)+f(x)f(y)=f(xy)+(y+1)f(x)+(x+1)f(y) $ $P(0,0) \implies {f(0)}^2=2f(0) $ if $ f(0)=2 $ from $P(x,0) \implies f(x)=x+2$ which is not a solution so $f(0)=0$ Put $P(1,-1) \implies f(1)f(-1)=3f(-1)$ if $ f(1)=3 $ $P(x-1,1) \implies f(x)=3x $ which is a solution so we have $f(-1)=0$ $P(-1,-1) \implies f(-2)=f(1) $ $P(-2,1) \implies f(-1)=f(-2)(3-f(1))+(-1)f(1) $ so $ 0=f(1)(2-f(1)) $ if f(1)=0 we get $P(x,-1) \implies f(x-1)=f(-x)$ $P(x-1,1) \implies f(x)=3f(x-1)=3f(-x)$ so from $ f(x)=3f(-x)$ we get $ f(x)=3f(-x)=3(3f(-(-x)))=9f(x) \implies f(x)=0$ which is a solution if f(1)=2 we get $P(x,-1) \implies f(x-1)=f(-x)$ $P(x-1,1) \implies f(x)=f(x-1)+2x=f(-x)+2x$ so $f(x)=f(-x)+2x$ $P(x,-x) \implies f(0)+f(x)f(-x)=f(-x^2)+(x+1)f(-x)+(-x+1)f(x)$ $0+f(x)(f(x)-2x)=(f(x^2)-2x^2)+(x+1)(f(x)-2x)+(-x+1)f(x) \implies f(x)^2-2xf(x)=f(x^2)-2x^2+xf(x)-2x^2+f(x)-2x-xf(x)+f(x)$ $f(x)^2=f(x^2)-4x^2+2f(x)+2xf(x)-2x$ $P(x,x) \implies f(2x)+ {f(x)}^2=f(x^2)+2(x+1)f(x)$ $f(2x)+(f(x^2)-4x^2+2f(x)+2xf(x)-2x)=f(x^2)+2(x+1)f(x)$ $f(2x)=4x^2+2x$ or $f(x)=x^2+x$ which also shows to be a solution

FantasyLover wrote: Rewriting the functional equation in terms of $g$, we have $g(x)g(y)=g(xy)+g(x)+g(y)-g(x+y)+2xy$. Just a comment: if we let $h(x) = g(x) - 1$, then $(h(x) + 1)(h(y) + 1) + h(x+y) + 1 = h(xy) + 1 + h(x) + 1 + h(y) + 1 + 2xy$, or $h(x)h(y) + h(x+y) = h(xy) + 2xy + 1$, which is exactly 2005 A4.

Let $P(x,y)$ be the assertion into the problem statement. $P(0,0) \implies f(0)^2=2f(0)$ so either $f(0)=0$ or $f(0)=2.$ Case 1: $f(0)=2.$ $P(x,0)\implies 3f(x)=2+f(x)+2x+2\implies f(x)=x+2$. Plugging this into the problem statement yields that it is not a solution. Case 2: $f(0)=0.$ $P(-1,1)\implies f(1)f(-1)=3f(-1)$ so $(f(1)-3)f(-1)=0$. Either $f(1)=3$ or $f(-1)=0.$ Case 2.1: $f(1)=3.$ $P(x-1,1)\implies \boxed{f(x)=3x}$ and plugging this in indeed works. Case 2.2: $f(-1)=0.$ $P(2,-1)\implies f(1)=f(-2).$ $P(-2,1)\implies f(-2)f(1)=3f(-2)-f(1).$ From above, $f(1)=f(-2)$ so it becomes $f(1)^2=2f(1)$ so either $f(1)=0$ or $f(1)=2$. Case 2.2.1: $f(1)=0$. $P(x,-1)\implies f(x-1)=f(-x). P(x-1,1)\implies f(x)=3f(x-1)$. Substituting $f(x-1)$ from first into the second yields $$f(x)=3f(-x).$$Let $x=x$ and $x=-x$ into the past function to get $f(x)=3f(-x)$ and $f(-x)=3f(x).$ Substituting the $f(-x)$ from the second into the first yields $f(x)=3f(-x)=3(3f(x))=9f(x)\implies \boxed{f(x)=0}.$ We can easily check that this function works. Case 2.2.2: $f(1)=2.$ $P(x-1,1)\implies f(x)=f(x-1)+2x$. $P(x,-1)\implies f(x-1)=f(-x)$. Combining these yields $$f(x)=f(-x)+2x.$$We can easily compute $f(2)=6$ by $P(1,1)$. $P(x,2)=f(2x)=f(x+2)+3f(x)-6x-6=f(x+1)+2(x+2)+3f(x)-6x-6=f(x)+2(x+1)+2(x+2)+3f(x)-6x-6=4f(x)-2x$, where we used $f(x)=f(x-1)+2x$ twice. $P(x,x)=f(2x)+(f(x))^2=f(x^2)+2xf(x)+2f(x)=4f(x)-2x+(f(x))^2+2xf(x)+2f(x)$, where we simplified $f(2x).$ $P(x,-x)=f(x)f(-x)=f(-x^2)+(x+1)f(-x)+(-x+1)f(x)$. Using $f(x)=f(-x)+2x, $ we have $f(-x^2)=f(x^2)-2x^2$ and $f(-x)=f(x)-2x.$ Substituting these in yields $f(x)^2-2xf(x)=f(x^2)-4x^2-2x+2f(x).$ Subtracting this from the $P(x,x)$ line gives $\boxed{f(x)=x^2+x}$ and we can easily check that this function works. Hence, the three solutions are $f(x)=0, f(x)=3x,$ and $f(x)=x^2+x$.