Ways to solve in integers $\frac{2x^2+5y^2}{xy-14}=11 $

Question

Consider the diophantine equation $$\frac{2x^2+5y^2}{xy-14}=11.$$ I have successfully found all its integer solutions, but in view of different equations, I was wondering if there are other approaches. I'll post mine as an answer.

Answer 1

Rewrite the equation as $$2x^2+5y^2=11xy-154 \\ 2x^2+5y^2-11xy=-154 $$ to note that decomposing $-11xy$: $$ 2x^2-xy+5y^2-10xy=-154 $$allows us to factor the LHS: $$ x ( 2x -y) - 5y (2x-y)=-154 \\ (x-5y)(2x-y)=-154.$$ Hence, as long as $154=2\cdot7\cdot11$, by the fundamental theorem of arithmetic, we just have a few systems to care about:$$\begin{cases} x-5y=\pm2 \\ 2x-y=\mp77 \\ \end{cases} \tag{1}$$ $$\begin{cases} x-5y=\pm14 \\ 2x-y=\mp11 \\ \end{cases} \tag{2}$$ $$\begin{cases} x-5y=\pm22 \\ 2x-y=\mp7 \\ \end{cases} \tag{3}$$ $$\begin{cases} x-5y=\pm154 \\ 2x-y=\mp1 \\ \end{cases} \tag{4}$$ $$\begin{cases} x-5y=\pm1 \\ 2x-y=\mp154 \\ \end{cases} \tag{5}$$ $$\begin{cases} x-5y=\pm7 \\ 2x-y=\mp22 \\ \end{cases} \tag{6}$$ $$\begin{cases} x-5y=\pm11 \\ 2x-y=\mp14 \\ \end{cases} \tag{7}$$ $$\begin{cases} x-5y=\pm77 \\ 2x-y=\mp2 \\ \end{cases} \tag{8}$$ which yield $(x,y)=(\pm9,\pm4),(\pm13,\pm4),(-43,-9)$.