11. Determine all functions \( f: \mathbf{N} \rightarrow \mathbf{N} \), such that
\[
f(a+b)=f(a)+f(b)+f(c)+f(d)
\]

for all non-negative integers \( a, b, c, d \) satisfying \( 2 a b=c^{2}+d^{2} \).
11. The function to be found is $f(n)=k n^{2}(k \in \mathbf{N})$.

Let $f$ be the function that satisfies the conditions.
Let $(a, b, c, d)=(n, n, n, n)$, then
$$
f(2 n)=4 f(n)
$$

holds for all non-negative integers $n$. In particular,
$$
\begin{array}{l}
f(0)=0, f(2)=4 k(k=f(1)) . \\
\text { Let }(a, b, c, d)=\left(n^{2}, 1, n, n\right), \\
(a, b, c, d)=\left(n^{2}, 2,2 n, 0\right), \\
(a, b, c, d)=\left(n^{2}+1,1, n+1, n-1\right) .
\end{array}
$$

Substituting the conditions respectively, we get
$$
\begin{array}{l}
f\left(n^{2}+1\right)=f\left(n^{2}\right)+k+2 f(n), \\
f\left(n^{2}+2\right)=f\left(n^{2}\right)+4 k+f(2 n) \\
=f\left(n^{2}\right)+4 k+4 f(n), \\
f\left(n^{2}+2\right) \\
=f\left(n^{2}+1\right)+k+f(n+1)+f(n-1) .
\end{array}
$$
(1) + (3) - (2) gives
$$
f(n+1)=2 f(n)-f(n-1)+2 k \text {. }
$$

Using induction on $n$, it is clear that
$$
f(n)=k n^{2}(k \in \mathbf{N}) .
$$