Say we have the quadratic Diophantine equation :
$4x^2+9y^2=100$
I'm not really sure what the usual method to find integer solutions to these equations are , but one way I thought of was the following :
we know $4x^2\geq 0 \Rightarrow 9y^2 \leq 100 \Rightarrow y^2\leq 11.111 \Rightarrow -3.333 \geq y \leq 3.333$, so we know $y=-3,-2,-1,0,1,2$ or $3$.
We must also have the condition that $x^2=\tfrac{100-9y^2}{4}$ is an integer, and the square of an integer.
giving
$y=0, x=5$
$y=0, x=-5$
$y=2, x=4$
$y=2, x=-4$
$y=-2, x=4$
$y=-2, x=-4$
these pairs making up the set of integer solutions.
My questions are , firstly is this an okay method ? Secondly is there a better method ? After all if we had a very large number on the R.H.S. this trial and error method would be very impractical , I feel there must certainly be a more efficient one.
Edit: Particularly I would like to know is there a better method than this one for the general quadratic Diophantine equation , $ax^2+by^2=c$