Problem: Given a2+b2=c2 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,
Squaring a number does not change its congruence mod 2.
Therefore c is even
Case 2: a is even, b is even. Similar to above
even2+even2=c2⟹c is even
One of a and b is odd, the other is even
Without loss of generality, we label a as odd, and b as even
Therefore c is odd
We have exhausted every possible case, and each shows a+b+c is even. QED
Is there a proof that doesn’t rely on case by case analysis?
Can the above be written in a simpler way?
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