I'm pondering the following claim regarding special cases of the Pell equation.
Conjecture: For every pair of distinct primes $p$ and $q$, there exist integers $a$ and $b$, and a non-square integer $d$, such that $(ap)^2-d(bq)^2=1$.
Is this a known theorem? Or are there known or easy counterexamples?
You can, in fact, do it with $b=1$, noting that this amounts to looking for a non-square integer of the form
$$d={(ap-1)(ap+1)\over q^2}$$
To find such a $d$, note that $\gcd(p,q)=1$ implies there is a number $a_0$ such that $a_0p\equiv1$ mod $q^2$. If we now take $a$ of the form $a_0+kq^2$, we see that $ap+1=(a_0p+1)+kpq^2$, which means, by Dirichlet's theorem for primes in airthmetic progressions, we can pick $k$ to make $ap+1=gP$, where where $g=\gcd(a_0p+1,pq^2)$ and $P$ is an arbitrarily large prime, which clearly cannot divide $ap-1$. For this choice, we have $P\mid d$ but $P^2\not\mid d$, so $d$ is not a square.
Note, we did not use the assumption that $p$ and $q$ are primes, just the (clearly necessary) condition $\gcd(p,q)=1$.