Prove that the sum of pythagorean triples is always even

Question 1

Problem: Given $a^2 + b^2 = c^2$ show $a + b + c$ is always even

My Attempt, Case by case analysis:

Case 1: a is odd, b is odd. From the first equation,

$odd^2 + odd^2 = c^2$

$odd + odd = c^2 \implies c^2 = even$

Squaring a number does not change its congruence mod 2.

Therefore c is even

$a + b + c = odd + odd + even = even$

Case 2: a is even, b is even. Similar to above

$even^2 + even^2 = c^2 \implies c$ is even

$a + b + c = even + even + even = even$

Case 3:
One of a and b is odd, the other is even
Without loss of generality, we label a as odd, and b as even

$odd^2 + even^2 = c^2 \implies odd + even = c^2 = odd$

Therefore c is odd

$a + b + c = odd + even + odd = even$

We have exhausted every possible case, and each shows $a + b + c$ is even. QED

Follow Up:
Is there a proof that doesn’t rely on case by case analysis?
Can the above be written in a simpler way?

Question 2

Note that $x^2\equiv x\pmod 2$ and thus $a^2+b^2=c^2$ implies $a+b+c\equiv a^2+b^2+c^2\equiv 2c^2\equiv 0\pmod 2$

Answer