Let $p$ and $q$ be prime numbers of the form $4k+3$. Suppose that there exist integers $x$ and $y$ such that $x^2-pqy^2=1$. Prove that there exist positive integers $a$ and $b$ such that $|pa^2-qb^2|=1$.
Problem
Source: Kürschák József Competition 2022/2
Tags: number theory, prime numbers, square
TheMathBob
08.10.2022 13:53
I think that it must be $x$ and $y$ non-zero, and $p\neq q$, because otherwise take $p=q=3$
TheMathBob
08.10.2022 14:22
If $p \neq q$, then
$$\left(\frac{p}{q}\right)\left(\frac{q}{p}\right) = (-1)^{\frac{p-1}{2}\frac{q-1}{2}}=-1,$$so $q$ is a quadratic residue modulo $p$ or $p$ is a QR$\pmod q$.
wlog let $\left(\frac{q}{p}\right)=-1$, then $\left(\frac{-q^{-1}}{p}\right)=\left(\frac{-q}{p}\right)=1$.
So there exists $t$ and $k$ such that $1+qt^2 = kp$, what's left is to prove that we can have such $t$, that $k$ is a square.
oVlad
09.10.2022 11:59
Kinda cute. As @TheMathBob said, we should assume that $x$ and $y$ are non-negative (but $p\neq q$ is not needed). Let $(x,y)$ be the pair satisfying $x^2-pqy^2=1$ with minimal $|xy|>0$. Note that $pqy^2=(x-1)(x+1)$. After a modulo $4$ analysis, it follows that $x-1=2u^2\delta$ and $x+1=2v^2(pq/\delta)$ for some divisor $\delta$ of $pq$ and integers $u,v$ satisfying $2uv=y$. Therefore, \[v^2\cdot\frac{pq}{\delta}-u^2\cdot\delta=1.\]If $\delta\in\{p,q\}$ then we get the desired identity. If $\delta=pq$ then as $|uv|=|y|/2<|xy|$ we reach a contradiction. Finally, if $\delta=1$ then $pq\mid u^2+1$ which is impossible, as $p\equiv q\equiv 3\bmod{4}$, so the proof is finished.