-1
$\begingroup$

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 $x=y=z=0$.

$\endgroup$
  • 1
    $\begingroup$ 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
3
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.