Answer
See below.
Work Step by Step
1. Use the method of contradiction, suppose the opposite is true:
both $a$ and $b$ are odd.
2. Let $a=2m+1$ and $b=2n+1$ where $m,n$ are integers, we have $c^2=a^2+b^2$ must be even, that is $c$ is even.
3. Let $c=2k$, we have $(2k)^2=(2m+1)^2+(2n+1)^2$ or
$4k^2=4(m^2+n^2)+4(m+n)+2$
4. We can see that the left side is divisible by 4 while the right side is not (with a remainder of 2), thus we reached a contradiction.
5. We can conclude that the statement in the exercise is true.
