How to solve $x^2+y^2=pz^2$ in $x,y,z\in\mathbb{N}$ if $p$ is such a prime that $p\equiv 3 \pmod 4$?

Is the following proof OK?

For such a prime we have lemma: $$p\mid a^2+b^2 \implies p\mid a\;\; {\rm and}\;\; p\mid b$$

Let's go to the problem:

Since $p\mid x^2+y^2$ we get $p\mid x$ and $p\mid y$ so $x=px'$ and $y=py'$ and now we have $$p^2x'^2+p^2y'^2 = pz^2\implies p\mid z $$ so $z=pz'$ and we get $$x'^2+y'^2=pz'^2$$ but this is the same equation as before with $x'$ smaller then $x$ and so on. So we can repeat this infinite times but this is impossible. So $x=y=z=0$.

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    $\begingroup$ I like to do this first: prove Lemma: that if there is any integer solution, not all zero, then there is such a solution with $\gcd(x,y,z) = 1.$ Part II: if $x^2 + y^2 - p z^2 \equiv 0 \pmod{p^2,}$ then $x,y,z$ are all divisible by $p.$ Thus, any solution, not all zero, has $x,y,z$ all divisible by $p$ and hence the gcd is not 1. ..... This places the "infinite descent" as the first item proved rather than the last. $\endgroup$ – Will Jagy Oct 4 '18 at 21:33
  • $\begingroup$ where was I: it also places the mod p stuff as a single finite check $\pmod {p^2}$ $\endgroup$ – Will Jagy Oct 4 '18 at 21:37
  • $\begingroup$ @DougM, actually, the claim does hold for $p \equiv 3 \pmod 4.$ The presentation above is a little brief, the OP could have said that was a lemma in some particular book. Anyway, it follows from Legendre symbol $(-1|p) = -1$ $\endgroup$ – Will Jagy Oct 4 '18 at 21:40
  • $\begingroup$ For an "alternative-proof": this $p$ is irredicuble in the UFD $\mathbb{Z}[i]$. So, if the two sides of the equation were nonzero, then the order of $p$ in $x^2 + y^2 = (x+iy) (x-iy)$ would be even whereas the order of $p$ in $p z^2$ would be odd, giving a contradiction. $\endgroup$ – Daniel Schepler Oct 4 '18 at 21:56
  • $\begingroup$ Note that your lemma can be proved as follows: $p\mid a^2+b^2 = (a+bi)(a-bi)$. But $p$ is prime in $\mathbb{Z}[i]$, so $p\mid a+bi$ or $p\mid a-bi$. In either case, this forces $p\mid a$ and $p\mid b$. $\endgroup$ – rogerl Oct 5 '18 at 0:54

Well, I am not sure either way about your lemma. The thing is though, there is a fairly short proof of the statement you are trying to show, that does not assume the lemma [which, if this were a homework question, would be what I think the instructor would want you to do].

The proof I had in mind:

If $x^2+y^2 = pz^2$, then we can assume that not all of $x,y,z$ are even [make sure you see why]. Consider two cases:

Case 1: $x,y$ odd. Then $x^2+y^2 \equiv 2$ mod 4, which implies that $pz^2 \equiv 2$ mod 4 which implies $z^2 \equiv 2$ mod 4 as $p \equiv 3$ mod 4. This is impossible as the only squares mod 4 are 0 or 1.

Case 2: $x$ odd, $y$ even. Then $x^2+y^2 \equiv 1$ mod 4 [why?] which implies $pz^2 \equiv 1$ mod 4, but as $p \equiv 3$ mod 4 implies $z^2 \equiv 3$ mod 4. This cannot be either, as the only squares mod 4 are 0 or 1.

Case 3: $x,y$ even. Then $pz^2$ must be even implying $z^2$ must be even, which contradicts not all of $x,y,z$ even.

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    $\begingroup$ yes, $x^2 + y^2 - p z^2 \equiv x^2 + y^2 + z^2 \pmod 4$ $\endgroup$ – Will Jagy Oct 4 '18 at 21:42
  • $\begingroup$ Yes @WillJagy that is a more concise way to say what I was getting at. Nice! $\endgroup$ – Mike Oct 4 '18 at 21:45
  • $\begingroup$ For an indefinite (integer coefficient) ternary quadratic form, either it really is possible to solve $f(x,y,z) = 0$ or there are an even number of primes, each of which can be used to prove impossibility. For this question, the primes are $2,3.$ This follows from something called the product formula for the Hilbert Norm Residue symbol. It is in Cassels, Rational Quadratic Forms. store.doverpublications.com/0486466701.html $\endgroup$ – Will Jagy Oct 4 '18 at 21:49
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    $\begingroup$ I will have to take your word for that @WillJagy. My knowledge of number theory is pretty elementary....I have heard of Legendre symbols but that is about it for me $\endgroup$ – Mike Oct 4 '18 at 21:54

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