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I am trying to find out how to solve $ax^2 - by^2 + cx - dy + e = 0$ to get integer solutions, failing this the rational solutions.

Thanks!

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When write the equation, the solutions will be determined by the solutions of certain equations Pell. –  individ Jun 10 at 5:25

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Solutions of the equation:

$ax^2-by^2+cx-dy+q=0$

you can record if the root of the whole: $k=\sqrt{(c-d)^2-4q(a-b)}$

Then using the solutions of the equation Pell: $p^2-abs^2=\pm1$

Then the formula of the solution, you can write:

$x=\frac{\pm1}{2(a-b)}(((d-c)\pm{k})p^2+2(bk\mp(bc-ad))ps+b(a(d+c)-2bc\pm{ak})s^2)$

$y=\frac{\pm1}{2(a-b)}(((d-c)\pm{k})p^2+2(ak\mp(bc-ad))ps-a(b(d+c)-2ad\mp{bk})s^2)$

If the root is a need to find out if this is equivalent to the quadratic form in which the root of the whole. This is usually accomplished this replacement: $x$ in such number $x+ty$

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Forgot to say. The characters inside the brackets do not depend on the sign of the Pell equation. It depends only before $\pm{1}$ –  individ Jun 10 at 13:21

Hint: Change it to $ax^2+cx+e=by^2+dy$ and complete the square on both sides. You will be left with a constant on one side. Put the squares on the other side from the positive constant and factor the difference of squares. For integer solutions, there are not many factorizations of the constant. Rational solutions are harder.

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No. This is not a solution. It is necessary in General to solve the equation. –  individ Jun 10 at 4:40

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