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The Diophantine equation of the form a$x^2$ – b$y^2$ = $c^2$ with ab is not a perfect square in Z has infinitely solutions in N, provided by a particular non-trivial solution in set of N.

I have racked my brains trying to think why ab not a perfect square should invalidate the proof, but can't think why. I have many books on number theory, but none have an equation like this.

If any one can help me in this aspect...I am so thankful to them.

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    @Daniel, OK, that clears things up. For a minute there I thought we had a proof of the inconsistency of mathematics!2012-03-21

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The choice of having $c^2$ on the right hand side is irrelevant, any nonzero number gives the same conclusions.

Let $ g = \gcd(a,b). $ By unique factorization, with $ \gcd \left( \frac{a}{g}, \frac{b}{g} \right) = 1, $ the product being a square gives $ a = g \alpha^2, \; \; b = g \beta^2, $ and let us take $g, \alpha, \beta > 0.$ So your equation becomes $ c^2 = a x^2 - b y^2 = g (\alpha^2 x^2 - \beta^2 y^2) = g (\alpha x - \beta y) (\alpha x + \beta y). $ Now, either $\alpha x, \; \beta y$ have the same sign or opposite. With $c \neq 0$ we get get one factor at least $1$ in absolute value, so then $ |\alpha x| + | \beta y| \leq \frac{c^2}{g}, $ so $ | x| \leq \frac{c^2}{g \alpha} $ and $ | y| \leq \frac{c^2}{g \beta}, $ giving finiteness of the set of solutions.

MEANWHILE, if $ab$ is not a perfect square, there are infinitely many solutions to the Pell equation $ u^2 - a b v^2 = 1. $ This makes infinitely many different solutions if there are any, because $ a (ux + b v y)^2 - b (avx + uy)^2 = a x^2 - b y^2. $

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I'm not sure what you are asking. If you are asking for an example where $ab$ is a perfect square and the equation doesn't have infinitely many solutions, perhaps the simplest example is that with $a=b=1$.