I've written down the following proof:

Given positive integers $x,y,z,k$ we have: $$[1]\;\;\;\;x^2+y^2+z^2=k^2, \rightarrow x^2+y^2=(k-z)(k+z)$$ We now let $k=z+1$, thus: $$x^2+y^2=2z+1$$ Let $z$ be even i.e. $z=2z'$, then:$$[2]\;\;\;\;x^2+y^2=4z'+1$$ We know that there exist an infinite amount of primes of the form $4m+1$, therefore there are infinitely many $z'$s such that $4z'+1$ is prime.

By Fermat's theorem in additive number theory we know that if $p \equiv 1 \pmod4$ then $p=u^2+v^2$ for some positive integer $u$ and $v$, since for every prime of the form $4z'+1 \equiv 1 \pmod4$, there are infinitely many solutions to $[2]$ therefore there are infinitely many solutions to $[1]$.

I know this proof is probably an "overkill" for the question and that I should prove the two statements I used (infinitely many primes and F.Theorem), but I think the proof is nonetheless correct right? Also, how can I prove that there are infinitely many solutions to $[1]$ such that $(x,y,z)=1$ ?

  • 2
    $\begingroup$ I guess a proof would be that: $x=1, y=2k$ and $z=2k^2$ does the work since: $$1+(2k)^2+(2k^2)^2=1+4k^2+4k^4=(1+2k^2)^2$$ which satisfies our conditions, but this is a sort of "parametrization" $\endgroup$ – Spasoje Durovic Feb 9 '19 at 15:06
  • $\begingroup$ See en.wikipedia.org/wiki/Pythagorean_quadruple $\endgroup$ – Robert Z Feb 9 '19 at 15:09

After thinking about it for a while I noticed that, if:$$4z'+1=x^2+y^2$$ Then $x$ and $y$ must have opposite parity, not only that, it must be the case that $(x,y)=1$, suppose to the contrary, $(x,y)=k$ for some $k\ge2$, then:$$x^2+y^2=(kx')^2+(ky')^2=k^2(y'^2+x'^2)$$ thus $k^2\;|\;x^2+y^2=4z'+1$ and that implies that $4z'+1$ is not prime, which is a contraddiction thus $(x,y)=1$ and therefore $(x,y,z)=1$


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