Solution

Prove that if $a$, $b$, $c$ is a primitive Pythagorean triple, $a$ and $b$ can't both be even.


We can use an indirect argument here.

Assume that both $a$ and $b$ are even. Then, for some integers $n_1$ and $n_2$, we must have $a=2n_1$ and $b=2n_2$.

Now consider whether $c^2$ can be even or odd: $$
\begin{align}c^2 &= a^2 + b^2 \\
&= (2n_1)^2 + (2n_2)^2 \\
&= 4n_1^2 + 4n_2^2 \\
&= 2(2n_1^2 + 2n_2^2) \\
&= 2n_3 \textrm{ for some integer $n_3$.}
\end{align}$$

So $c^2$ must be even. Consequently, $c$ itself must be even. (Clearly, $c$ can't be odd, as the square of an odd number is odd.) Notice, we now have $a$, $b$, and $c$ are all even, and hence, all divisible by $2$.

Recall, however, a "primitive" Pythagorean triple can't have any common divisors (other than 1). So we have a contradiction! Thus, we reject our original assumption that $a$ and $b$ are even. The opposite must instead be true: $a$ and $b$ can't both be even.