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Prove that the sum of the squares of two odd integers cannot be the square of an integer.

My method:

Assume to the contrary that the sum of the squares of two odd integers can be the square of an integer. Suppose that $x, y, z \in \mathbb{Z}$ such that $x^2 + y^2 = z^2$, and $x$ and $y$ are odd. Let $x = 2m + 1$ and $y = 2n + 1$. Hence, $x^2 + y^2$ = $(2m + 1)^2 + (2n + 1)^2$ $$= 4m^2 + 4m + 1 + 4n^2 + 4n + 1$$ $$= 4(m^2 + n^2) + 4(m + n) + 2$$ $$= 2[2(m^2 + n^2) + 2(m + n) + 1]$$ Since $2(m^2 + n^2) + 2(m + n) + 1$ is odd it shows that the sum of the squares of two odd integers cannot be the square of an integer.

This is what I have so far but I think it needs some work.

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    $\begingroup$ You’ve shown that the sum of the squares of two odd integers is of the form $4\ell+2$. Now show that there is no integer whose square has this form. The square of an odd integer has the form $4\ell+1$, and the square of an even integer is divisible by ... ? $\endgroup$ – Brian M. Scott May 1 '16 at 19:52
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    $\begingroup$ This is quite fine. For completeness you might want to add that an even square must be the square of an even number / divisible by 4.-- Anyone more experienced might have remembered that odd squares are $\equiv 1\pmod 8$, hence the sum of two such is $\equiv 2\pmod 8$, which cannot be square. Your argument boils down to working $\pmod 4$, which is in fact sufficent $\endgroup$ – Hagen von Eitzen May 1 '16 at 19:54
  • $\begingroup$ I want to edit this because the second-last sentence reads "the sum of two odd integers." Is this a mistake? $\endgroup$ – ahorn May 1 '16 at 20:00
  • $\begingroup$ @ahorn I just edited it. Was what I fixed what you were talking about? $\endgroup$ – Matt May 1 '16 at 20:06
  • $\begingroup$ @Matt yes. I wasn't 100% sure, so I wanted to check. $\endgroup$ – ahorn May 1 '16 at 20:08
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Let $a=2n+1$, $b=2m+1$. Then $a^2 + b^2=4n^2 + 4n +4m^2 +4m+2$. This is divisible by $2$, a prime number, but not by $4=2^2$. Hence it cannot be the square of an integer.

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You had a great start but you should not make this last factorisation.

$$4(m^2 + n^2) + 4(m + n) + 2=4(m^2+n^2+m+n)+2=2 \pmod 4$$

But a square can't be equal to $2 \pmod 4$.

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  • $\begingroup$ Since I have $z^2$ = $x^2 + y^2$ would I have to show a case where $z^2$ is odd? $\endgroup$ – Matt May 1 '16 at 20:14
  • $\begingroup$ If $z^2$ is odd it is the same, $z^2 \neq 2 \pmod 4$ $\endgroup$ – Jennifer May 1 '16 at 20:18
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    $\begingroup$ and $z^2$ can't be odd, because $x$ odd $\iff x^2$ odd, so $x^2$ and $y^2$ are odd, so $x^2+y^2=z^2$ is even. $\endgroup$ – Jennifer May 1 '16 at 20:22
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Here's a quick method, not unrelated to your approach or to the other answers here.

The squares mod $4$ are $0$ and $1$ (can be verified easily by checking all four). Odd numbers are congruent to $1$ or $3$ mod $4$ and these each have square congruent to $1$ mod $4$. Hence the sum of two odd squares is congruent to $2$ mod $4$ which isn't a square.

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