wangsacl wrote: Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that for all $x,y\in\mathbb{R}$, then \[f(x+f(y))=f(x-f(y))+4xf(y)\] where $\mathbb{R}$ denote the set of real numbers. Let $P(x,y)$ be the assertion $f(x+f(y))=f(x-f(y))+4xf(y)$ The function $f(x)=0$ $\forall x$ is a solution. Let us from now look for non all zero solutions. Let $u$ such that $f(u)\ne 0$ $P(\frac x{8f(u)},u)$ $\implies$ $f(\frac x{8f(u)}+f(u))=f(\frac x{8f(u)}-f(u))+\frac x2$ And so $x=2f(\frac x{8f(u)}+f(u))-2f(\frac x{8f(u)}-f(u))$ and so any real $x$ may be written as $2f(y)-2f(z)$ for some $y,z$ $P(-f(z),z)$ $\implies$ $f(0)=f(-2f(z))-4f(z)^2$ $P(f(y)-2f(z),y)$ $\implies$ $f(2f(y)-2f(z))=f(-2f(z))+4f(y)^2-8f(y)f(z)$ Subtracting these two lines, we get : $f(2f(y)-2f(z))=(2f(y)-2f(z))^2+f(0)$ And so $f(x)=x^2+f(0)$ $\forall x$ which indeed is a solution. Hence the two solutions : $f(x)=0$ $\forall x$ $f(x)=x^2+a$ $\forall x$

Notice the uncanny resemblance (if not utter equivalence) with Problem 3 at the 2007 Balkan Mathematical Olympiad. See http://www.artofproblemsolving.com/Forum/viewtopic.php?p=827178&sid=0d0afe7c258254c48c003fe03fe58e44#p827178.

Can anybody post other problems of first day? Thanks!!!

$f(x+f(y)) = f(x-f(y))+4xf(y) $. (1) First note that $f(x)=0$ for any $x$ is solution to the given equation. Suppose there exist some $u$ such that $f(u)$ doesn't equal to $0$. Then, $4xf(u)$ can get equal to any real number, but by (1) it's the difference of two f-s. It means that any real can be expressed as a $f(a)-f(b)$. (2) Denote $f(0)=c$. Than, setting $x=f(y)$ in (1), we get: $f(2f(y))=(2f(y))^{2}+c $ (3). Furthermore, setting $f(x)$ instead of $x$ in (1), $f(f(x)-f(y)=f(f(x)+f(y))-4f(x)f(y)$. So, as the right hand side is symmetric, $f(f(x)-f(y))=f(f(y)-f(x))$. Hence, $f$ is odd, because $f(x)-f(y)$ gets any real value by (2). Denote by $A$ the set of all $a$, satisfying $f(a)=a^{2}+c$. From (3), $2f(x)$ belong to $A$ and as $f$ is odd, also $-2f(x)$ belong to that set. From (1), it's clear that if $a-b=2f(x)$ for some $x,a,b$, than $f(a)-f(b)=a^{2}-b^{2}$. So, if $a$ is in $A$, than so does $a+2f(x)$. From the last fact, as $-2f(a)$ is in$A$ for any $a$, than for any $b$, $-2f(a)+2f(b)=2(f(b)-f(a))$ is in $A$. But $f(b)-f(a)$ gets any real value, as well as $2(f(b)-f(a))$. So, $A=R$. It means, from the definition of $A$, that $f(x)=x^{2}+c$ holds for any $x$. Finally, the only solutions are $f(x)=0$ and $f(x)=x^{2}+c$. P.S. The problem is not easy, but the approach is quite well known

lasha wrote: P.S. The problem is not easy, but the approach is quite well known What about this approach is well known? (I don't see what the main idea, that is said to be well known is?)

The main idea is that the expression $f(a)-f(b)$ may take on any real value. Such a step is certainly not easy to motivate. Indeed, it is the key step of IMO 1999 Q6 (so it's quite well known). However, after seeing a solution using this method, the method is easy to mimic for similar questions.

If we put $f(x)=x^2 + g(x)$, then we get LHS=RHS for all $x,y$. Does that mean anything?

If you do that, you get $g(x+y^2 + g(y)) = g(x-y^2 - g(y))$. It remains for you to prove that this functional equation has as only solutions $g(x)=-x^2$ (leading to $f \equiv 0$), and $g(x) = c$ (constant, leading to $f(x) = x^2+c$).

MatjazL wrote: lasha wrote: P.S. The problem is not easy, but the approach is quite well known What about this approach is well known? (I don't see what the main idea, that is said to be well known is?) Sorry for a late response, I've just came across it Anyway, jjax anwered to you correctly. That approach has been used in IMO 1999/6 (But not only, I've solved several problems with that, just don't remember the sources).

waver123 wrote: If we put $f(x)=x^2 + g(x)$, then we get LHS=RHS for all $x,y$. Does that mean anything? solution with $g(x)$ is pretty good, you may see in http://www.artofproblemsolving.com/Forum/viewtopic.php?p=827178&sid=f14af484eddd9e2dc421aec7daa5ed21#p827178 for solution by Anto

I don't know if it's good but i think to create the function $g(x)=f(x)-x^2$ we can write $4xf(y)=(x+f(y))^2-(x-f(y))^2$ and we have that $g(x+f(y))=g(x-f(y)) $ and we can put y=0 and note that f(0)=a and we have that $f(k)=k^2+g(f(0))=k^2+f(a)-a^2$

$P(x+f(y),z)$ yields $f(x+f(y)+f(z))=f(x+f(y)-f(z))+4(x+f(y))f(z)$ $P(x-f(y),z)$ yields $f(x-f(y)+f(z))=f(x-f(y)-f(z))+4(x-f(y))f(z)$ adding the two equations yields $f(x+f(y)+f(z))-f(x+f(y)-f(z))+f(x-f(y)+f(z))-f(x-f(y)-f(z))=8xf(z)$ changing $y$ and $z$ yields $f(x+f(y)+f(z))+f(x+f(y)-f(z))-f(x-f(y)+f(z))-f(x-f(y)-f(z))=8xf(y)$ subtracting the two equations yields $f(x+f(y)-f(z))-f(x-(f(y)-f(z)))=4x(f(y)-f(z))$ since $f(a)-f(b)$ runs through the entire real numbers(when $f$ is nontrivial) we can write $$f(x+y)-f(x-y)=4xy$$thus $f(x)=x^2+c$, $f=0$

Let P(x,y) be the assertion f(x-f(y)) = f(x) -f(y)

liimr wrote: waver123 wrote: If we put $f(x)=x^2 + g(x)$, then we get LHS=RHS for all $x,y$. Does that mean anything? solution with $g(x)$ is pretty good, you may see in http://www.artofproblemsolving.com/Forum/viewtopic.php?p=827178&sid=f14af484eddd9e2dc421aec7daa5ed21#p827178 for solution by Anto @liimr Are you sure that this solution can be used for this problem? We can get $g$ to be periodic but then in the $x=y+T$ equation $g(y)$ does not cancel and so we cannot work with $y = (-z-2T^2)/4T$ (or a similar expression) here.