General way of solve $ax^2+by+c=0$ For example ,the diophantine equation
$$x^2+1=25y$$
we can solve this by finding particular solution $(x,y)=(7,2)$
and using this , we can get general solution.
My question is 
"To solve $ax^2+by+c=0$ $(a, b, c \in \mathbb{Z})$ we must find particular solution or 
not?"

Since x and y have a difference like $d$,we can have a single variable quadratic equation by transformation.
If $x=y ±d$ we have:
$a(y ±d)^2+by+c=0$
$ay^2 +(b ±2ad)y  ±ad^2+c=0$
$\Delta=(b ±2ad)^2-4a(c ±ad^2)>=0$
Limitation of value of $\Delta$ gives us a short range of numbers to try for solution. For example we solve $x^2+1=25y$:
We consider two cases:
$|x|>|y|$; so we can transform equation as follows:
Let $x= y+d$ , we have:
$(y+d)^2-25y+1=0$
$y^2+(2d-25)y+d^2+1=0$
$\Delta=-100 d +621$
For a solution in set of real numbers we must have:
$\Delta=-100 a +621>=0$  ⇒ $100a<=621$  ⇒$a<6$ .
Now we have a small range to try, for example :
$a=5$ ⇒$\Delta=121$ ⇒ $y=2$ and $y=13$
$y=2  ⇒ x=7$
$y=13 ⇒ x=18$.
Now suppose $|x|<|y|$, we have:
$(y-d)^2+1-25y=0$
$y^2+(-2a-25)y+a^2+1=0$
Which gives $y=2, x=7$ and $y=41, x=32$
For a general formula for x and y, the homogeneous equation $x\times x-25y=0$  gives $x=25$ and $y=1$  and we have:
$x=7$ as a single solution so the equation for x is $x=25t +7$.
Where t ∈ Z. For y we put x in equation:   
$(25t+7)^2+1=25y$  which gives $y=25t^2+14t+2$.
Hence there are infinitely many solutions for this equation.
