Let $a,b,c$ be integers with $a$ positive. Is there a criterion for $ax^2+bx+c$ with $x$ integer being a square number? That is, is there a criterion for the equation $ax^2+bx+c=y^2$ has an integer solution $(x,y)$?
For example if $(a,b,c)=(1,0,0)$ then there will be infinitely many solutions, but if $(a,b,c)=(2,0,0)$ then there will be no solution.
Also I wonder that if there is a solution, then there are infinitely many.
P.S. There is a similar question number of solutions of a quadric diophantine equation here, but I'm not sure that it is equivalent to my question. We can complete the square $ax^2+bx+c=a(x+\frac{b}{2a})^2+(c-\frac{b^2}{4a})$ and substitute $x'=x+\frac{b}{2a}$, but $\frac{b}{2a}$ and $\frac{b^2}{4a}$ may not be integers.
We may write:
$A=ax^2+bx+c=(x+\frac{b-\sqrt{\Delta}}{ 2a})(x-\frac{-b-\sqrt{\Delta}}{ 2a})=(x+\frac {b}{2a}-\frac{\sqrt {\Delta}}{2a})(x+\frac {b}{2a}+\frac{\sqrt {\Delta}}{2a})$
$A$ can be a perfect square if $\Delta=0$, so we must have:
$\Delta= b^2-4ac=0$
For example:
$a=1, c=4\rightarrow b^2= 4^2\rightarrow b=\pm 4$
and we get :
$x^2\pm4x+4=(x\pm2)^2$
But a cannot be equal to c unless it is a perfect square:
$a=c=2\rightarrow b^2=4^2\rightarrow b=\pm4$
and we get:
$2x^2\pm4x+2=2(x\pm1)^2$
Which is not a perfect square,but for $a=c=4$ we get:
$b^2=4\times 4^2=8^2\rightarrow b=\pm 8$
and we have:
$4x^2\pm8x+4=4(x+1)^2=[2(x+1)]^2$
Generally for $a=1$ and $c=4^{2n-1}$ we have:
$b^2=\pm 4^{2n}\rightarrow b=\pm 4^n$
and the equation becomes:
$x^2\pm 4^n x+4^{2n-1}=(x+2^{2n-1})^2$
Examples:
$n=2\rightarrow 2n-1 =3\rightarrow c=4^3, b^2=4\times 4^3=4^4\rightarrow b=\pm 4^2$
and equation becomes:
$x^2\pm 4^2 x+4^3=(x\pm 2 ^3)^2$
similarly we can get:
$n=3\rightarrow x^2\pm 4^3x+ 4^5=(x\pm 2^5)^2$
$n=4\rightarrow x^2\pm 4^4x+ 4^7=(x\pm 2^7)^2$
etc...
In relation $\Delta=b^2-4ac$ , $a$ and $c$ are cyclic , so may also have:
$4^{2n-1}x^2\pm4^n x+1=(2^{2n-1}x\pm 1)^2$
Or $a=c=4^{2n-1}$ which gives:
$4^{2n-1} x^2\pm 4^n x+4^{2n-1}=(2^{2n-1}x\pm 2^{2n-1})^2$