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How do you prove that if each each of two natural numbers $a$ and $b$ is a sum of two squares then $ab$ is also a sum of two squares?

I'm pretty sure that this is a true statement but I don't know how to prove this.

Bill Dubuque
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Chris
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    $(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2$ – Will Jagy Jul 14 '14 at 18:18
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    It will be easier to appreciate the identity if you split any number who is a sum of two squares as a product of gaussian integers. $$\begin{align} \big(a^2+b^2\big)\big(c^2+d^2\big) &= \big((a+bi)(a-bi)\big)\big((c+di)(c-di)\big)\ &=\big((a+bi)(c+di)\big)\big((a-bi)(c-di)\big)\ &= \big((ac-bd)+(ad+bc)i\big)\big((ac-bd)-(ad+bc)i\big)\ &= (ac-bd)^2 + (ad+bc)^2 \end{align} $$ – achille hui Jul 14 '14 at 18:56

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For any integer $a,b,c,d$, $$(a^2+b^2)\cdot (c^2+d^2)=(ac-bd)^2+(ad+bc)^2.$$ This can be derived from the norm $N(\mathcal{ab})=N(\mathcal{a})N(\mathcal{b})$ for integral ideal $\mathcal{a},\mathcal{b}\in \mathbb{Q(i)}$.

user164341
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