Question

Solve in positive integers
$2xy-4x^2+12x-5y=11$.

Answer 1

Expressing $y$ in terms of $x$, we have
$y=\frac{4x^2-12x+11}{2x-5}=2x-1+\frac{6}{2x-5}$.
In order that $y$ may be an integer $\frac{6}{2x-5}$ must be an integer; hence $2x-5$ must be equal to $\pm 1$, or $\pm 2$, or $\pm 3$, or $\pm 6$.
The cases $\pm 2$, $\pm 6$ may clearly be rejected;

Hence the admissible values of $x$ are obtained from $2x-5=\pm 1$, $2x-5=\pm 3$;
whence the values of $x$ are $3,2,4,1$.
Taking these values in succession we obtain the solutions
$x=3$, $y=11$; $x=2$, $y=-3$; $x=4$, $y=9$; $x=1$, $y=-1$;
and therefore the admissible solutions are
$x=3$, $y=11$; $x=4$, $y=9$.