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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.

Matt
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    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 01 '16 at 19:52
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    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 01 '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 01 '16 at 20:00
  • @ahorn I just edited it. Was what I fixed what you were talking about? – Matt May 01 '16 at 20:06
  • @Matt yes. I wasn't 100% sure, so I wanted to check. – ahorn May 01 '16 at 20:08
  • @BrianM.Scott why cant we write $4l+2$ = $2k$ ? so question would be then that number of form $2k$ cannot be square of integer. – Sophie Clad Mar 22 '23 at 12:18
  • @SophieClad: But every square of an even number is of the form $2k$: $2^2=2\cdot 2$, $4^2=2\cdot 8$, $6^2=2\cdot 18$, and so on. The point here is that these squares are all of the form $4k$, not $4k+2$. – Brian M. Scott Mar 22 '23 at 20:40

5 Answers5

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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.

user1043065
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You had a great start but you should not make this last factorisation.

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

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

amWhy
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Bérénice
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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.

john
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just think base2: a and b, being odd numbers will look like $$ xxxx1_2$$ square of a and b will be odd and also look like $$ xxxx1_2 $$ sum of $$xxxx1_2 + xxxx1_2 = xxx10_2 $$

square of even number $$xxxx0_2$$ will look like $$xxx00_2$$

so digits on position '2' will be always 0 for square of even number but for sum of squares of odd numbers will be 1:

$$ xxxx10_2 \neq xxx00_2 $$

Michal Kupczyk
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The exercise 5.28 of the book Mathemathical Proof is exactly this. I was trying to prove it, and I was about to ask about it, but I find your question.

The difference is here (I want to know if what I have done is valid):

$$x^{2}+y^{2} = 4m^{2}+4m+4n^{2}+4n+2 = 2(2m^{2}+2m+2n^{2}+2n+1) = z^{2}$$

Since $2m^{2}+2m+2n^{2}+2n+1$ is an integer, $2|z^{2}$ and so $2|z$.

But $z = \sqrt{2(2m^{2}+2m+2n^{2}+2n+1)}$, so $z$ is irrational (because $\sqrt{2}$ is irrational). So we have a contradiction.

P.S.: Please, be kind. I have received a lot of negative points these last few days, and I am just trying to do my best. Thank you.

David
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  • We have $2|z^2$ but $4$ does not divide $z^2$. Contradiction. Why should we mention about the irrationality of $z$? – Bob Dobbs Aug 27 '23 at 11:27
  • I don't know. I thought that if $z$ is irrational, 2 does not divide $z$. So, since $2|z^{2}$ and so $2|z$, we have a contradiction. I am a begginer, I am starting a degree in maths this year at university. – David Aug 27 '23 at 14:39