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I would like to know all integer solutions of the Diophantine equation $x^2-ay^2-bz^2+abw^2=1$ where $a,b$ are fixed positive integers. If you know the answer, I appreciate your reply. (To be more specific, $a,b$ are such that $x^2-ay^2-bz^2+abw^2=0$ has no solution)

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  • $\begingroup$ Decisions can be recorded through the solution of the equivalent equation Pell. The formula is somewhat cumbersome. You look like you need? $\endgroup$ – individ Mar 27 '17 at 4:20
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The simplest approach I think is.

$$x^2-ay^2-bz^2+abw^2=1$$

To make the change. $y=kS$ ; $z=tS$ ; $w=pS$

Reduced to a Pell equation. That factor was not square. And using his solution we write the answer.

$$x^2-(ak^2+bt^2-abp^2)S^2=1$$

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  • $\begingroup$ This approach helps a lot, thank you. $\endgroup$ – Mohammadreza Bidar Mar 27 '17 at 21:05

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