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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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What is a Dio-Equation? Please consider spending more time to write up your question. Your interest is what can convince people that you have thought about that question. – user21436 Mar 21 '12 at 4:40
$ab$ could be a perfect square, and you could still have infinitely many solutions. – Daniel Montealegre Mar 21 '12 at 4:45
@Daniel, that's not in accord with Will's answer. – Gerry Myerson Mar 21 '12 at 5:53
@GerryMyerson Sorry I didnt know that $c$ was fixed, the wording was rather confusing. I was thinking if $a=b=1$, then you could have $(x,y,c)$ being a pythagorean triple – Daniel Montealegre Mar 21 '12 at 5:59
@Daniel, OK, that clears things up. For a minute there I thought we had a proof of the inconsistency of mathematics! – Gerry Myerson Mar 21 '12 at 22:12

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 $a=b=1$.

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