Problem: "Let m and n be integers such that each can be expressed as the sum of
two perfect squares. Show that mn has this property as well"
Proof: Let $m =a^2+b^2$ , $n=c^2+d^2$ ,where $a, b , c , d$ are integers. Now consider the product $$z:=(a+bi)(c+di)$$ Note that $$|z|=|(a+bi)||(c+di)|=\sqrt{(a^2+b^2)(c^2+d^2)}$$Therefore $mn = |z|^2$. But $|z|^2$ will be a sum of squares of integers, since $\Re (z)$ and $\Im(z)$
are integers.
Everything is clear except for the last statement, "But $|z|^2$ will be a sum of squares of integers, since $\Re (z)$ and $\Im(z)$
are integers."
Best Answer
Note that$$\lvert z\rvert^2=\operatorname{Re}^2z+\operatorname{Im}^2z.$$On the other hand,$$\operatorname{Re}z=ac-bd\in\mathbb Z\text{ and }\operatorname{Im}z=ad+bc\in\mathbb Z.$$