Solution

$\require{AMSsymbols}$Prove that if $a$, $b$, $c$ is a primitive Pythagorean triple, $a$ and $b$ can't both be odd.


We can use an indirect argument here.

Assume that both $a$ and $b$ are odd. Then, for some integers $n_1$ and $n_2$, we must have $a=2n_1+1$ and $b=2n_2+1$.

Now consider $c^2$: $$
\begin{align}c^2 &= a^2 + b^2 \\
&= (2n_1 + 1)^2 + (2n_2 + 1)^2 \\
&= 4n_1^2 + 4n_1 + 1 + 4n_2^2 + 4n_2 + 1 \\
&= 4n_1^2 + 4n_1 + 4n_2^2 + 4n_2 + 2 \\
&= 4(n_1^2 + n_1 + n_2^2 + n_2) + 2 \\
&= 4n_3 + 2 \textrm{ for some integer $n_3$}
\end{align}$$

Notice, this tells us two things:

  • $c^2$ must be even as $4n_3+2 = 2(2n_3 + 1)$
  • $4 \nmid c^2$, as the division would result in a remainder of 2.

However, if $c^2$ is even, then $c$ itself must be even. (Clearly, $c$ can't be odd, as the square of an odd number is odd.) Thus, $c=2k$ for some integer $k$. This, in turn, implies that $c^2 = (2k)^2 = 4k^2$. So, is must be true that $4 \mid c^2$. This contradicts our earlier conclusion that $4 \nmid c^2$. So we are forced to reject our original assumption that $a$ and $b$ are both odd. The opposite must instead be true: $a$ and $b$ can't both be odd.