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What would be a parametric characterization of all relatively prime solutions in positive integers to $x^2 + y^2 = 2z^2$?

The hint I got was: Show there are integers $a$ and $b$ for which $x = a+b$ and $y = a−b$.

What I thought

I thought that I could do this by first showing that z must be odd, and then x and y are odd, but I don't seem to be able to do this.

Martin Sleziak
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Sandi kae
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1 Answers1

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Existence of such integers $a,b$ is equivalent to saying that $x,y$ have the same parity (do you see why?). If $x,y$ had opposite parity, then $x^2+y^2$ would be odd, while $2z^2$ is even.

Now if we have such $a,b$ then we can substitute to get $(a+b)^2+(a-b)^2=2z^2$. Simple algebra gives us that this is equivalent to $a^2+b^2=z^2$. I suspect that now you know the parametrization of solutions to this equation (you also have to show that $a,b,z$ are relatively prime; I leave it to you).

As you have asked for this, here is how to continue: because $(a,b,z)$ is a primitive Pythagorean triple, we have, for some integers $m,n$, that $a=m^2-n^2,b=2mn,z=m^2+n^2$. Now we have $x=a+b=m^2+2mn-n^2,y=a-b=m^2-2mn-n^2$. So the complete parametrization is:

$$(m^2+2mn-n^2,m^2-2mn-n^2,m^2+n^2)$$

Wojowu
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  • I understood that part, but I had trouble going on from this. – Sandi kae May 14 '15 at 21:11
  • Does this help for this?: The pythagorean triples are multiples of the form $(x-y,x+y,2z)=(m^2-n^2,2mn,m^2+n^2)$ implies $(x,y,z)=k\cdot (\tfrac{m^2+2mn-n^2}{2},\tfrac{2mn-m^2+n^2}{2},\tfrac{m^2+n^2}{2})$. – Sandi kae May 14 '15 at 21:14
  • We have arrived at a Pythagorean triple $(a,b,z)$. Use parametrization here to find what form $a,b,z$ have, and to find $x,y$ recall that $x=a+b,y=a-b$. – Wojowu May 14 '15 at 21:32
  • Am I right in the above comment? – Sandi kae May 14 '15 at 21:33
  • I don't know how you have got this first triple you mention. – Wojowu May 14 '15 at 21:35
  • Can you show it if you can. Im really having trouble with this – Sandi kae May 14 '15 at 21:36
  • @Sandikae I have completed the solution for you. – Wojowu May 15 '15 at 05:19
  • There is a minor bug here: $m^2-2mn-n^2$ might be negative, and instead of $a = m^2-n^2$ and $b=2mn$ you might need $a = 2mn$ and $b = m^2-n^2$. (As it stands, your parametrization only covers the case when $8 \mid x-y$; in particular, it misses the solution $x = 7, y = 17, z = 13$.) – darij grinberg May 16 '16 at 21:00