SOLUTION: Prove:the sum of the squares of two odd integers cannot be a perfect square, i.e, if x and y are both odd, then there is no perfect square z^2 such that x^2+y^2=z^2.

Algebra ->  Proofs -> SOLUTION: Prove:the sum of the squares of two odd integers cannot be a perfect square, i.e, if x and y are both odd, then there is no perfect square z^2 such that x^2+y^2=z^2.      Log On


   

Question 974893: Prove:the sum of the squares of two odd integers cannot be a perfect square, i.e, if x and y are both odd, then there is no perfect square z^2 such that x^2+y^2=z^2.
Answer by amarjeeth123(569) About Me  (Show Source):
You can put this solution on YOUR website!
let the two odd integers be (2m+1) and (2n+1).
The sum of their squares is (2m+1)^2+(2n+1)^2
=4m^2+4m+1+4n^2+4n+1
=4(m^2+n^2)+4m+4n+2
=4(m^2+n^2+2mn-2mn)+4(m+n)+2
=4(m+n)^2+4(m+n)-8mn+2
This is a quadratic in (m+n).
For this expression to be a perfect square discriminant=0
16-16(-8mn+2)=0
-16(-8mn+2)=-16
-8mn+2=1
-8mn=-1
n=1/8m
But m and n both have to be integers, which means it is impossible for n=1/8m.
Therefore there is no way for the sum of squares of two odd integers to also be a perfect square.