If $3\omega^2-2n\omega\psi+\psi^2+1=0$ has a rational solution then it has an integer solution?

Question

I want to prove this as a middle step in this thing and it occurred to me that someone might know a proof or a counterexample as a canned fact. In fact if this is true then it finishes the proof that $x^2-(n^2-3)y^2=-1$ is solvable if and only if $n=\frac{3p^2+q^2-r^2\pm 2r^2}{2pr}$ for integers $p,q,r$.

I called the given solution $(\omega_0,\psi_0)$ then $(\omega_0+t,\psi_0+kt)$ is also a solution for all rational $k$ and

$$t=\frac{2n\omega_0k+2n\psi_0-6\omega_0-2k\psi_0}{k^2-2nk+3}$$

But this is quite ugly and I haven't managed to extract an integral solution from it.

Answer 1

Here is a counterexample.

Let $n=52$ to get the equation $$ 3x^2-104xy^2+y^2+1=0 $$ We may verify that there is a rational solution $(x,y)=(1/10,1/10)$.

Now rearranging and letting $w=y-52x$, this can be written as negative Pell equation: $$ w^2-2701x^2=-1 $$ If there is an integral solution $(x,y)$ to the original equation, we would also get an integral solution to the negative Pell equation. However it is known that this negative Pell equation has no solutions. For example it's not in this solvable list from OEIS.

You may also try to find the fundamental units of $\mathbb Q[\sqrt{2701}]$ and show that there are none with norm $-1$. I've not tried working it out, only checked against a commonly used calculator here.