proof of a statement about the Diophantine equation $ax^2-by^2=c^2$ 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.
 A: 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
$$ a (ux + b v y)^2 - b (avx + uy)^2 = a x^2 - b y^2.     $$
A: 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 $a=b=1$. 
