# Solving the equation $x^2-7y^2=-3$ over integers

I'd like to solve the following Pell equation: $$x^2-7y^2=-3$$ Where $x$ and $y$ are integers. I applied the usual procedure, which avoids continued fractions:

The two minimal positive integer solutions are $(x_0,y_0)=(2,1)$ and $(x_1,y_1)=(5,2)$, thus the minimal rational solution of $x^2-7y^2=1$ should be $(p,q)=\left(\frac{4}{3},\frac{1}{3}\right)$. My script (it is in german so I don't link it here) tells me, that in this case, every pair of solutions is given by: $$x_{n+1}=\frac{4}{3}x_{n}+7\cdot\frac{1}{3}y_n \\ y_{n+1}=\frac{1}{3}x_{n}+\frac{4}{3}y_n$$ If we proceed further, we can find that this gives: $$x_n=\frac{a_n}{3^n} \space\text{where}\space a_0=2,\space a_1=15,\space a_{n+1}=8a_n-9a_{n-1} \\ y_n=\frac{b_n}{3^n} \space\text{where}\space b_0=1,\space b_1=6,\space b_{n+1}=8b_n-9b_{n-1}$$ But if we take these equations modulo $9$, we see that $(2,1)$ and $(5,2)$ are the only integer solution, but there surely is also $(37,14)$. Where did I go wrong? Every answer will be appreciated, but I'm not used to the approach with continued fractions, so preferably I would like to see an answer avoiding this.

EDIT:

My main question is:

Where is my fault? Or is my script wrong?