Question

Solve in positive integers the equation
$x^2-4xy+6y^2-2x-20y=29$.

Answer 1

Solving as a quadratic in $x$, we have
$x=2y+1\pm \sqrt{30+24y-2y^2}$.
But $30+24y-2y^2=102-2{(y-6)}^2$; hence ${(y-6)}^2$ cannot be greater than $51$. By trial we find that the expression under the radical becomes a perfect square when ${(y-6)}^2=1$ or $49$; thus the positive integral values of $y$ are $5,7,13$.
When $y=5$, $x=21$ or $1$; when $y=7$, $x=25$ or $5$; when $y=13$, $x=29$ or $25$.