Can You Prove that if x2+y2 is even, then x+y is even.

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Today we prove that if x squared plus y squared is even, then x plus y is even. #MathProofs #DiscreteMath #Proofs
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[@brownsugarissupreme]

Oh that's really smart. The method I immediately thought of is to use contraposition

[@pulse.7]

you can also prove by case. For x^2 + y^2 to be even, then either they are both even or both odd, and then from there its quite easy to show the answer

[@michaeleissenberg637]

can i prove by contraposition?
if x²+y² is even, then x+y is even, by contraposition we have: if x+y is odd then x²+y² is odd. Then we assume x+y is odd is true, so x = 2n+1 (with n \in Z) and y = 2m+1 (with m \in Z). x²+y² = (2n+1)²+(2m+1)² = 4m²+4m+1+4n²+4n+1 = 4(m²+m+n²+n)+2 then we divide the equation by 2 so we have : 2(m²+m+n²+n) + 1 and we know (m²+m+n²+n) is an integer. So we've proven by contraposition that if x²+y² is even, then x+y is even ?? i don't know, please help me.

[@Trevtutor]

There’s an error in your first step of the assumption that x+y is odd. That doesn’t imply that both x and y are odd, because that would make x+y even. It means one of x and y is odd and one of x and y is even. That’ll get you on track.

[@michaeleissenberg637]

thanks a lot!@@Trevtutor