# Supposed $a,b \in \mathbb{Z}$. If $ab$ is odd, then $a^{2} + b^{2}$ is even.

Supposed $a,b \in \mathbb{Z}$. If $ab$ is odd, then $a^{2} + b^{2}$ is even.

I'm stuck on the best way to get this started. My thinking is that I could use cases. i.e.

• Case 1: a is even and b is odd
• Case 2: a is odd and b is even
• Case 3: a is odd and b is odd

Would this be my best approach? Or is there an easier way to look at it? Thanks.

If $ab$ is odd then $a$ and $b$ must be both odd and then so are $a^2$ and $b^2$. But $a^2+b^2$ is the sum of two odd numbers, so it must be even.
$ab$ is odd iff a and b are both odd. That means $a^2$ and $b^2$ are both odd. The sum of two odd numbers is even. Therefore, $a^2+b^2$ must be even if $ab$ is odd.
More generally note: $\ ab(a^2+b^2)\,$ is even since
${\rm mod}\ 2\!:\ x^2\equiv x\,$ $\,\Rightarrow\,ab(a^2+b^2)\equiv ab(a+b)\equiv ab+ab\equiv 2ab\equiv 0$