Prove that $S = \{a^2 + b^2: a,b \in \Bbb N\}$ is closed under multiplication. Is it true? 
Can you prove or disprove this?
$S = \{a^2 + b^2: a,b \in \Bbb N\}$ is closed under multiplication.

Write

$$(a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ad+bc)^2.$$

This comes from me knowing that $a^2+b^2=(a+bi)(a-bi) = |a+bi|^2|$. Here $i=\sqrt{-1}$. Then I just multiply $(a+bi)(c+di) = (ac-bd)+i(ad+bc)$, and since

$$|zw|^2=|z|^2|w|^2$$

So that means we know that $(ac-bd)^2+(ad+bc)^2=|(ac-bd)+i(ad+bc)|^2=|a+bi|^2|c+di|^2=(a^2+b^2)(c^2+d^2)$.
