Prove the sum of squares of two odd integers can't be a perfect square

  • Thread starter kolley
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  • #1
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Homework Statement



x^2+y^2=z^2

Homework Equations





The Attempt at a Solution



assume to the contrary that two odd numbers squared can be perfect squares. Then,
x=2j+1 y=2k+1

(2j+1)^2 +(2k+1)^2=z^2
4j^2 +4j+1+4k^2+4k+1
=4j^2+4k^2+4j+4k+2=z^2
=2[2(j^2+K^2+j+k)+1)]=2s
the book goes on to produce a contradiction having to do with whether s is odd or even. Can someone please walk me through the rest of this proof to the end in detail, because the explanation in my book is very poor. Thank you.
How do I go on to show that this is a contradiction. The book says
 

Answers and Replies

  • #2
UD1
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Note that the sum of the square of two odd integers (2m+1) and (2n+1) is X=4(n^2+m^2+n+m)+2. Clearly, X is congruent to 2 modulo 4. But k^2 can only give remainder 0 or 1 modulo 4.
 

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