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,

odd2+odd2=c2

odd+odd=c2⟹c2=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

even2+even2=c2⟹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 evenodd2+even2=c2⟹odd+even=c2=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?

**Answer**

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

**Attribution***Source : Link , Question Author : spyr03 , Answer Author : Ennar*