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

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.

• 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 ... ? – Brian M. Scott May 1 '16 at 19:52
• 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 – Hagen von Eitzen May 1 '16 at 19:54
• I want to edit this because the second-last sentence reads "the sum of two odd integers." Is this a mistake? – ahorn May 1 '16 at 20:00
• @ahorn I just edited it. Was what I fixed what you were talking about? – Matt May 1 '16 at 20:06
• @Matt yes. I wasn't 100% sure, so I wanted to check. – ahorn May 1 '16 at 20:08

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.

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 $2 \pmod 4$.

• Since I have $z^2$ = $x^2 + y^2$ would I have to show a case where $z^2$ is odd? – Matt May 1 '16 at 20:14
• If $z^2$ is odd it is the same, $z^2 \neq 2 \pmod 4$ – Jennifer May 1 '16 at 20:18
• 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. – Jennifer May 1 '16 at 20:22

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.