10. (20 points) Define the function $f: \mathbf{Z} \rightarrow \mathbf{Z}$, for any $x, y \in \mathbf{Z}$, we have
$$
f\left(x^{2}-3 y^{2}\right)+f\left(x^{2}+y^{2}\right)=2(x+y) f(x-y) \text {. }
$$

If $n>0$, then $f(n)>0$, and $f(2015) f(2016)$ is a perfect square. Find the minimum value of $f(1)+f(2)$.
10. Replace $y$ with $-y$ in the equation, we get
$$
f\left(x^{2}-3 y^{2}\right)+f\left(x^{2}+y^{2}\right)=2(x-y) f(x+y) \text {. }
$$

Combining with the original equation, we have
$$
(x-y) f(x+y)=(x+y) f(x-y) \text {. }
$$

This indicates that, when $a, b \in \mathbf{Z}-\{0\}$, and $a \equiv b(\bmod 2)$, $\frac{f(a)}{a}=\frac{f(b)}{b}$, meaning there exists a constant $\alpha=f(1) \in \mathbf{Z}_{+}$, $\beta=f(2) \in \mathbf{Z}_{+}$, such that $f(n)$ is defined as follows:
$$
f(n)=\left\{\begin{array}{ll}
n \alpha, & 2 \nmid n ; \\
\frac{n}{2} \beta, & 2 \mid n .
\end{array}\right.
$$

Then $f(2015) f(2016)=2015 \alpha \times 1008 \beta$
$$
\Rightarrow \alpha \beta=5 \times 7 \times 13 \times 31 t^{2}\left(t \in \mathbf{Z}_{+}\right) \text {. }
$$

Thus, $(\alpha, \beta, t)=(5 \times 31,7 \times 13,1)$ determines the smallest $\alpha+\beta=246$.
Next, we prove: $\alpha+\beta \geqslant 246$.
When $t \geqslant 2$, by the basic inequality, we get
$$
\alpha+\beta \geqslant 2 \sqrt{\alpha \beta}>246 \text {; }
$$

When $t=1$, $\alpha+\beta=246$.
Finally, when $n$ is odd, define the function $f(n)=91 n$; when $n$ is even, define the function $f(n)=\frac{155}{2} n$.
It is easy to verify that $f(n)$ satisfies the condition
$$
f(1)+f(2)=246 \text {. }
$$