Find all real real functions that satisfy the following eqation $f(x^2)+f(2y^2)=[f(x+y)+f(y)][f(x-y)+f(y)]$ Find all real functions $f:\Bbb R\rightarrow\Bbb R$ so that $f(x^2)+f(2y^2)=[f(x+y)+f(y)][f(x-y)+f(y)]$, for all real numbers $x$ and $y$.
$f(x)=x^2$ is the only solution I think.
So far I have got:
$f(x^2)=[f(x)]^2+2f(x)f(0)+[f(0)]^2-f(0)$
$f(2y^2)=4[f(y)^2]-f(0)$
And by putting $x=y$ and $x=-y$ you get that $[f(x)-f(-x)][f(2x)-f(0)]=0$. If $f(2x)=f(0)$ then $f$ is a constant function of value $0$ or $1/2$, both work, if not than $f$ is an even function.
Source: 3rd European mathematical cup, senior category.
 A: Very partial solution.
I show that $f$ is even  when $f(0)=0$. I use a comment of @soulEater: in this case we have
$$(f(x+y)+f(x-y))(f(y)-f(-y))=0$$
Now suppose that there exists an $y_0$ such that $f(y_0)-f(-y_0)\not =0$, put $L=2y_0\not = 0$. We have $f(x+y_0)=-f(x-y_0)$ for all $x$, hence $f(x+L)=-f(x)$ and $f(x+2L)=f(x)$ for all $x$ (and $f$ is periodic).
We replace $x$ by $x+L$ in the original equation 
$$f((x+L)^2)+f(2y^2)=(-f(x+y)+f(y))(-f(x-y)+f(y))$$
We put $x=0$
$$f(L^2)+f(2y^2)=0$$
From here, we get $f(u)=c=-f(L^2)$ for all $u\geq 0$, and $f(u)=c$ for all $u$ by periodicity. But this contradict the existence of $y_0$. Hence we have $f(y)-f(-y)=0$ for all $y$, and $f$ is even.
EDIT
I suppose now that $\displaystyle f(0)=\frac{1}{2}$, and I suppose that $f$ is continuous.
I use again
$$(f(x+y)+f(x-y))(f(y)-f(-y))=2f(0)(f(y)-f(-y))$$
and a $y_0$ such that $f(y_0)-f(-y_0)\not =0$, and I put $L=2y_0\not = 0$. We have $f(x+y_0)=-f(x-y_0)+1$ for all $x$, hence $f(x+L)=-f(x)+1$ and $f(x+2L)=f(x)$ for all $x$ (and $f$ is periodic). Put $T=2L$.
From the original equation, replacing first $x$ by $x+T$, we get that the function $g(x)=f(x^2)$ has period $T$. Replacing $y$ by $y+T$, we get $f((\sqrt{2}y+T\sqrt{2})^2)=g(\sqrt{2}y+T\sqrt{2})=f(2y^2)=g(\sqrt{2}y)$, hence that for all $z$ we have $g(z+\sqrt{2}T)=g(z)$. Hence $g$ is periodic of period $T$ and period $\sqrt{2}T$, hence any number in $(\mathbb{Z}+\mathbb{Z}\sqrt{2})T$ is a period, i.e the set of period of $g$ is dense in $\mathbb{R}$. Now as $g$ is continuous, we get that $g$ is constant, hence $f$ is constant on $[0,+\infty[$, hence constant on $\mathbb{R}$ par periodicity. Again, this contradict the existence of $y_0$, so $f(y)-f(-y)=0$ for all $y$, and $f$ is even. 
A: There are three non-symmetries between the LHS and RHS of your equation: $x\to-x$
 , $y\to-y$, and $x^{2}\leftrightarrow2y^{2}$.
From $y\to-y$ after you subtract and make $x=0$ you get $f^{2}(y)=f^{2}(-y)$ for every y (see barto's comment).
From $x\to-x$ after you subtract and make $y=0$ you get $f(0)(f(x)-f(-x))=0$.
This implies that $f$ is always even, namely, if $f(0)\neq0$ then it is clear; otherwise $f$  is even from Kelenner's answer.
To be continued...
