When $a^2x^2+2bx+c$ is a perfect square?

Question

Is there any fast and easy method to find out $a^2x^2+2bx+c$ is a perfect square for some $x\in\mathbb{Z}$, i.e, when there is a $x,y\in\mathbb{Z}$ such that $$ a^2x^2+2bx+c=y^2 $$ where $a,b,c$ are given integers?

P.S.: My question has two parts:

First decide if it is perfect square or not

Second fast and easy method to find $x,y$.

Thanks!

Answer 1

I can't answer you full question but I may give you a way. Were can write $a^2x^2+2bx+c$=$$(ax)^2+2ax(\frac {b}{a})+(\frac {b}{a})^2+[c-(\frac {b}{a})^2]$$=$$(ax+\frac {b}{a})^2+[c-(\frac {b}{a})^2]$$ Thus if the second part i.e. $[c-(\frac {b}{a})^2]$ is equal to 0 then it is a perfect square. Then $ax+(\frac ba)^2$=$y$. So you can find the value of $x,y$. I am now still finding a good way. Thank you.

Answer 2

Given $a,b,c$ with $a\neq 0$, rewrite the equation in factored form

$$\left(a y-a^2x-b\right)\left(a y+a^2x+b\right)=a^2c-b^2$$

This leads to two linear equations

$$a y-a^2x-b=d_1$$

and

$$a y+a^2x+b=d_2$$

where $d_1, d_2$ are divisors of $a^2c-b^2$ such that $d_1 d_2=a^2c-b^2$