TITLE: 3xy + 14x + 17y + 71 = 0 need some advice
QUESTION [6 upvotes]: $$3xy + 14x + 17y + 71 = 0$$
Need to find both $x$ and $y$. If there was only one variable then this is easy problem.
Have tried:
$$\begin{align}3xy &= -14x - 17y - 71 \\
x &= \frac{-14x - 17y - 71}{3y}\end{align}$$
Then tried to put this expression everywhere instead of $x$ but it tooks forever to find both $x$ and $y$.
I don't even know how to get on right track.
Please give any advice. Thanks.

REPLY [4 votes]: This is another way to find the integer solutions.
Define $d=y-x$ and the equation becomes $3x(x+d)+14x+17(x+d)+71=0$
which boils down to $$3x^2+(3d+31)x+17d+71=0.$$
The quadratic formula gives the roots:$$6x=-(3d+31)\pm\sqrt{9d^2-18d+109},$$
and we want the discriminant, $9(d-1)^2+100 = (3(d-1))^2+10^2$ to be a square.
$d=1$ is obviously one candidate, leading to final values $x=-4, y=-3$.  
Otherwise we are just looking for Pythagorean triples. A quick look at a list of Pythagorean triples shows that there is just one triple with $10$ as one of the lower values: (10, 24, 26).
Thus $9(d-1)^2=24^2$, giving $d=9$ or $d=-7$, from which the appropriate values of $x$ and $y$ can be found.