Show that if $p$ and $q$ are primes $\equiv 3$(mod $4$) then at least one of the equations $px^{2}-qy^{2} = \pm 1$ is soluble in integers $x, y$.

Question

So far we have only talked about equations in the form $x^{2} - dy^{2} = \pm 1$ and I'm unsure how to handle the specific coefficients. We have normally found $\sqrt{d}$ and looked at the convergents of the continued fraction to find solutions. I was just wondering how to get started on this type of problem.

Answer 1

You have a primitive solution to $x^2-pqy^2=1$ ($x$, $y$ positive, $y$ minimal). Then $x$ is odd and $y$ is even (think modulo $4$). Also $$x^2-1=(x+1)(x-1)=pqy^2$$ so that $$\frac{x+1}2\frac{x-1}2=pq\left(\frac{y}{2}\right)^2.$$ As $(x\pm 1)/2$ are coprime integers then we have one of the following

1. $(x+1)/2=u^2$, $(x-1)/2=pqv^2$

2. $(x+1)/2=pu^2$, $(x-1)/2=qv^2$

3. $(x+1)/2=qu^2$, $(x-1)/2=pv^2$

4. $(x+1)/2=pqu^2$, $(x-1)/2=v^2$

We succeed with (2) and (3) (write $1=(x+1)/2-(x-1)/2$). We eliminate (1) by minimality to solution of Pell and (4) gives $v^2-pqu^2=-1$ which is impossible modulo $p$.