What's the best technique to find integer solutions for $x$ & $y$ to equations in the form $x^2-yx+ay+b = 0$ where $\space(a,b)\space$ are known integers (or even just to find the number of solutions that exist)?
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Add your attempts to your question thanks – user577215664 Aug 21 '20 at 11:17
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$x^2-yx+ay+b = 0 \implies (y - 2 a)^2 - (2 x - y)^2 = 4 (a^2 + b)$ – Dmitry Ezhov Aug 21 '20 at 11:40
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Let $u=x-a$. The equation becomes
$$uy=(u+a)^2+b=u(u+2a)+a^2+b$$
If $a^2+b\not=0$, we must have $u\mid a^2+b$, in which case there are only finitely many solutions. If $a^2+b=0$, the equation to solve is $uy=u(u+2a)$, with infinitely many solutions of the form $(u,y)=(0,y)$ and $(u,y)=(u,u+2a)$.
Barry Cipra
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The equation factors as follows: $a^2+b=(y-x-a)(x-a)$. Please refer to the comment above by @Dmitry Ezhov – Lozenges Aug 21 '20 at 12:16
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@Lozenges, arghh, another math error. I thought making a change of variable made it easier to see what's going on, but I see it also made it all too easy for me to make stupid mistakes! – Barry Cipra Aug 21 '20 at 14:35
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Give $\quad x^2-yx+ay+b = 0\quad$ we know we can get integer $(y)$ solutions for any integer $x\ne a$.
$$y =\frac {-b + x^2}{a - x} \land a \ne x$$
If we then solve for $x$, we can use any of the resulting $y$-values to get an $x$
$$x = \frac{y - \sqrt{y (y - 4 a) - 4 b}}{2}$$
but this is a round-robin thing. We have seen above that there are infinitely many solutions: $$a,b,y\in\mathbb{Z}\quad\land\quad x\in\mathbb{Z}-\{a\}$$
poetasis
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Equation given by "OP" is equal to:
$x^2+b=y(x-a)$ ---(1)
Equation $(1)$ has solution:
$x=3(2k-1)$
$y=(11k^2-10k+2)$
$a=6(k-1)$
$b=-3(k-1)^2$
For, $k=2$, we get:
$(a,b)=(6,-3)$ and
$(x,y)=(9,26)$
Sam
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