Proving even and odd pythagorean triples

Question

What is the most simplistic way to prove that, in a Pythagorean triple $(a,b,c)$ where $a^2 + b^2 = c^2$. If $c$ is even then so are both $a$ and $b$.

Is it necessary to use modular arithmetic or is there another way?

Answer 1

Hint:

every primitive Pythagorean triple $a,b,c$ can be generated by: $$ a=m^2-n^2 \qquad b=2mn \qquad c=m^2+n^2 $$ with $m-n$ odd and $m,n$ coprime.

Answer 2

I will try to keep this fairly straightforward.

---

Method: By contradiction.

Given: $c=2m$ ($m \in \mathbb{Z}$) i.e. $c$ is even, and $a^2+b^2=c^2$ (let Eq.1)

Case 1: Assume both $a$ and $b$ are odd. Let $a=2k+1$ and $b=2l+1$ for $k,l \in \mathbb{Z}$. Substitute $a,b$ in Eq. 1 and proceed to prove that the new equation cannot hold.

Case 2: Assume, wlog, $a=2k$ (even) and $b=2l+1$ (odd) for $k,l \in \mathbb{Z}$. Again, substitute in equation 1 and proceed to prove that the neww equation cannot hold.

Since, we have successfully contradicted both Case 1 and 2, the only possibility is that both $a$ and $b$ are even. Q.E.D.