, 2018

Let $a=f(0)$ and denote ($*$) as the given condition. Assume that $a \neq 0$, substituting $x=2a$ and $y=0$ in ($*$), we get
$$2a^{2}=\frac{(f(2a))^{2}}{2a}+a$$
or
$$4a-2=\left(\frac{f(2a)}{2a}\right)^{2}.$$
This follows that $4a-2$ is a perfect square which is a contradiction since there is no perfect square congruent to $2$ modulo $4$. Thus $a=0$, or $f(0)=0$. Now, substituting $y=0$ in ($*$), we get
$$x^{2} f(-x)=(f(x))^{2}, \quad \forall x \in \mathbb{Z}, x \neq 0$$
It follows that
$$x^{2} f(x)=(f(-x))^{2}, \quad \forall x \in \mathbb{Z}, x \neq 0$$
Therefore
$$(f(x))^{4}=x^{4}(f(-x))^{2}=x^{6} f(x), \quad \forall x \in \mathbb{Z}, x \neq 0$$
From this and $f(0)=0$, we get $f(x)=0$ or $f(x)=x^{2}$ for any integer $x$. Clearly, the function $f(x)=x^{2}$ satisfies the original equation. Let us consider now the case there exists some integer $c \neq 0$ such that $f(c)=0$. Replacing $y=c$ in ($*$), we get
$$c^{2} f(2x)=0, \quad \forall x \in \mathbb{Z}, x \neq 0$$
Thus, for any even integer $x$, we have $f(x)=0$. Now, if there exists some odd number $d$ such that $f(d)=d^{2}$, taking $x$ even and $y=d$ in ($*$), we get
$$d^{2} f\left(2x-d^{2}\right)=f\left(d^{3}\right), \quad \forall x \in \mathbb{Z}, x \neq 0, 2 \mid x$$
If $f\left(d^{3}\right) \neq 0$, then $f\left(2x-d^{2}\right) \neq 0$ and hence $f\left(d^{3}\right)=d^{6}, f\left(2x-d^{2}\right)= (2x-d^{2})^{2}$. Plug back into the above equation, we get a contradiction. Therefore, $f\left(d^{3}\right)=0$ and hence $f\left(2x-d^{2}\right)=0$ for any nonzero even integer $x$. It follows that $f(x)=0$ for any integer $x$ such that $x \equiv -1 \pmod{4}$ and $x \neq -d^{2}$.
In $x^{2} f(-x)=(f(x))^{2}$, taking $x \equiv -1 \pmod{4}, x \neq -d^{2}$, we get $f(-x)=0$. It follows that $f(x)=0$ for any integer $x$ such that $x \equiv 1 \pmod{4}$ and $x \neq d^{2}$. Thus $f(x)=0$ for any integer $x \neq \pm d^{2}$. However, we have $f(d)=d^{2} \neq 0$, so $d= \pm d^{2}$ or $d= \pm 1$. It follows that $d=d^{3}$ and so $f(d)=f\left(d^{3}\right)=0$, which is a contradiction. Therefore, in this case, we have $f(x)=0$ for any integer $x$.
In conclusion, the equation has two solution: $f(x)=0$ and $f(x)=x^{2}$.