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Is there a method that can solve all quadratic diophantine equations of the following type

$$X (X + a) = Y (Y + b)$$

where $a,b$ are given integers?

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  • $\begingroup$ I think they always have finitely many solutions unless $a=b$ or $a=-b$ and that the solutions are the 4 from making both sides zero plus at most a couple extra small solutions. $\endgroup$ – rain1 May 25 at 5:52
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$X (X + a) = Y (Y + b) \implies (2 X + a)^2 - (2 Y + b)^2 = a^2 - b^2$

Get finite set solutions of difference of squares $x^2 - y^2 = a^2 - b^2$ and check $X=\frac{x-a}{2}$ and $Y=\frac{y-b}{2}$ as integers.

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Above equation shown below:

$X (X + a) = Y (Y + b)$

Take, $[(a+b),(a-b)]=(4mp,4nq)$

$X=[(p-q)(n-m)]$

$Y=[(q-p)(m+n)]$

We get:

$a=2(mp+nq)$

$b=2(mp-nq)$

For, $(p,q,m,n)=(2,3,5,3)$ we get:

$(X,Y,a,b)=(2,8,38,2)$

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