, 2019

The required functions are $f(n) = k n^2$, where $k$ is a non-negative integer — these clearly satisfy the condition in the statement.

Conversely, let $f$ be a function satisfying the condition in the statement. Setting $(a, b, c, d) = (n, n, n, n)$ in the functional relation yields $f(2n) = 4f(n)$ for all $n$. In particular, $f(0) = 0$ and $f(2) = 4k$, where $k = f(1)$.

Setting successively $(a, b, c, d) = (n^2, 1, n, n)$, $(a, b, c, d) = (n^2, 2, 2n, 0)$ and $(a, b, c, d) = (n^2 + 1, 1, n + 1, n - 1)$ in the functional relation yields
$$\begin{align*} f(n^2 + 1) &= f(n^2) + k + 2f(n), \\ f(n^2 + 2) &= f(n^2) + 4k + f(2n) = f(n^2) + 4k + 4f(n), \\ f(n^2 + 2) &= f(n^2 + 1) + k + f(n + 1) + f(n - 1). \end{align*}$$
Subtraction of the second relation above from the sum of the other two yields
$f(n + 1) = 2f(n) - f(n - 1) + 2k$.

As in the first solution, consider a function $f$ satisfying the condition in the statement, and establish that $f(2n) = 4f(n)$, for all $n$; in particular, $f(0) = 0$.

We now show by induction on $m$ that $f(mn) = m^2 f(n)$ for all $n$. By the preceding, this is clearly the case if $m = 0, 1, 2$. For the induction step, let $m > 2$.

Since $\sqrt{m} < m$, if $m$ is a square, the conclusion follows by the induction hypothesis: $f(mn) = f(\sqrt{m}(\sqrt{m}n)) = m f(\sqrt{m} n) = m^2 f(n)$.

Otherwise, use Lagrange's four-square theorem to write $m = A + B$, where $A$ and $B$ are positive integers, each of which is a sum of two squares. Recall that if each factor of a product is a sum of two squares, then so is the product (use standard identities such as $(w^2 + x^2)(y^2 + z^2) = (wy + xz)^2 + (wz - xy)^2$ repeatedly), to write $2AB = (1^2 + 1^2)AB = C^2 + D^2$ for some non-negative integers $C$ and $D$. Clearly, $A < m$ and $B < m$; $C \le \sqrt{2AB} \le (A+B)/\sqrt{2} = m/\sqrt{2} < m$, and similarly, $D < m$. Setting $a = An, b = Bn, c = Cn$ and $d = Dn$ in the functional relation and applying the induction hypothesis completes the induction step:
$$\begin{aligned} f(mn) &= f((A+B)n) = f(An) + f(Bn) + f(Cn) + f(Dn) \\ &= A^2 f(n) + B^2 f(n) + C^2 f(n) + D^2 f(n) = (A^2 + B^2 + C^2 + D^2)f(n) \\ &= (A^2 + B^2 + 2AB)f(n) = (A+B)^2 f(n) = m^2 f(n), \end{aligned}$$
for all non-negative integers $n$.

Consequently, $f(mn) = m^2 f(n)$ for all $m$ and all $n$. Setting $n = 1$ and $k = f(1)$ concludes the proof.