I am relatively new at solving these kind of equation and was wondering if someone can help with a step by step for an odd formed Diophantine equation. The particular equation I am trying to solve is $-x^2 -2xy + 44y = 43$.
The integer solutions are not in general form but are
$${y = -286, x = 23}$$ $${y = -286, x = 549}$$ $${y = -46, x = 39}$$ $${y = -46, x = 53}$$ $${y = 2, x = 5}$$ $${y = 2, x = -9}$$ $${y = 242, x = -505}$$ $${y = 242, x = 21}$$
On
When you have a quadratic, the quadratic formula is your friend. Write it as $$x^2+2xy+43-44y=0\\x=\frac 12\left(-{2y}\pm\sqrt{4y^2-172+176y}\right)\\ =\frac 12\left(-{2y}\pm\sqrt{(2y+44)^2-1764}\right)$$ and you need to look for $y$s that will make the square root an integer. If you define $z=2y+44$ and require $z$ to be even you are looking at $$z^2-42^2=w^2\\(z+w)(z-w)=42^2$$ and look at the factorizations of $42^2$ into even numbers.
Hint: The equation is equivalent to $$ (x+2y+22)(22-x)=17\cdot31 $$