4. Let $a, b, c, d$ be integers, and $m=a^{2}+b^{2}, n=c^{2}+d^{2}$, then $m n$ can also be expressed as the sum of two squares of integers. Its form is: $m n=$ $\qquad$
4.
$$
\begin{aligned}
m n & =\left(a^{2}+b^{2}\right) \cdot\left(c^{2}+d\right)^{2} \\
& =(a c+b d)^{2}+(a d-b c)^{2} \\
& =(a c-b d)^{2}+(a d+b c)^{2}
\end{aligned}
$$

Filling out either of the two forms above is correct.
Note: From the solution, it is easy to derive the following conclusion:
If $a, b, m ; c, d, n$ are all Pythagorean triples, then $a c+b d,|a d-b c|, m n ;|a c-b d|, a d+b c, m n$ also form Pythagorean triples.