Estonian Mathematical Olympiad

Answer: $f(x)=0$ and $f(x)=x$.

Substituting $x=y=0$ to the equation, we get $(f(0){)}^2=0+2f(0)+(f(0){)}^2$. Simplifying this gives $f(0)=0$.

Substituting $y=0$ to the original equation, we get the equation $(f(x){)}^2=xf(x)+2f(0)+(f(0){)}^2$ which must be satisfied for all real $x$. Since $f(0)=0$, this simplifies to $f(x)(f(x)-x)=0$. Thus, for each $x$, either $f(x)=0$ or $f(x)=x$.

Assume there exists a real number $c\ne 0$ such that $f(c)=0$. Substituting $x=c$ to the original equation, we get $(f(c+y){)}^2=2f(cy)+(f(y){)}^2$, or $$2f(cy)=(f(c+y){)}^2-(f(y){)}^2.(1)$$ This is valid for any $y$. Let's analyse four cases based on whether $f(c+y)=0$ or $f(c+y)=c+y$ and whether $f(y)=0$ or $f(y)=y$.

1) If $f(c+y)=0$ and $f(y)=0$, then (1) simplifies to $f(cy)=0$.

2) If $f(c+y)=0$ and $f(y)=y$, then (1) gives $f(cy)=-\frac{y^2}{2}$. Assuming that $f(cy)=cy\ne 0$, we get $c=-\frac{y}{2}$ i.e. $y=-2c$.

3) If $f(c+y)=c+y$ and $f(y)=0$, then (1) gives us $2f(cy)=(c+y{)}^2$. Assuming that $f(cy)=cy$, we get $c^2+y^2=0$ from which $c=0$, contradiction.

4) If $f(c+y)=c+y$ and $f(y)=y$, then (1) gives us $2f(cy)=c^2+2cy$. Assuming that $f(cy)=cy$, we get $c^2=0$, contradiction.

Thus $f(cy)\ne 0$ can only be valid if $y=-2c$, i.e. $f(x)\ne 0$ can only be valid if $x=c\cdot (-2c)=-2c^2$. Thus it is possible to choose a real number $d$ such that $d\ne 0$, $f(d)=0$, and $|d|\ne |c|$. By replacing $d$ by $c$ we can analogously conclude that $f(x)\ne 0$ can only be valid when $x=-2d^2$. As $-2d^2\ne -2c^2$, $f(x)\ne 0$ cannot be valid for any $x\ne 0$. So $f(x)=0$ for all $x$.

Therefore the only suitable functions are $f(x)=0$ and $f(x)=x$.