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The question can be equated to the counting of solutions to the following diophantine equation: $$4a^4x^2\pm y^2=\Delta_f,$$ where $\Delta_f=b^2-4a^2c$ and plus\minus relate to considering when $f(x)=-\square$ or $\square$.

In the special case when $a=1$ more can be said. In particular, for counting negatives squares, there are known formulas for the $r_2(\Delta_f)$.

The question can be equated to the counting of solutions to the following diophantine equation: $$4a^4x^2\pm y^2=\Delta_f,$$ where $\Delta_f=b^2-4a^2c$ and plus\minus relate to considering when $f(x)=-\square$ or $\square$. This follows from the fact that if $\pm \square=a^2x_1^2+bx_1+c$, then substituting $x=x_1+\frac{b}{2a^2}$ and clearing the denominator we get the above equation.

In the special case when $a=1$ more can be said. In particular, for counting negatives squares, there are known formulas for the $r_2(\Delta_f)$.

The question can be equated to the counting of solutions to the following diophantine equation: $$4a^4x^2\pm y^2=\Delta_f,$$ where $\Delta_f=b^2-4a^2c$ and plus\minus relate to considering when $f(x)=-\square$ or $\square$. Thus we have a Ramanujan–Nagell type equation.

In the special case when $a=1$ more can be said. In particular, for counting negatives squares, there are known formulas for the $r_2(\Delta_f)$.

The question can be equated to the counting of solutions to the following diophantine equation: $$4a^4x^2\pm y^2=\Delta_f,$$ where $\Delta_f=b^2-4a^2c$ and plus\minus relate to considering when $f(x)=-\square$ or $\square$.

In the special case when $a=1$ more can be said. In particular, for counting negatives squares, there are known formulas for the $r_2(\Delta_f)$.

The following extremely special case may hint how to tackle the general question. If we assume that $a =1$, $f(x)=(x-\alpha_1)(x-\alpha_2)$ is an irreducible polynomial. Let it have a squarefree discriminant $\Delta_f$, such that none of the primes divisors of $\Delta_f$ are congruent to $3\pmod{4}$ (and thus $\alpha_i\in \mathbb{R}$). Then the number of $z \in (\alpha_1,\alpha_2)\cap\mathbb{Z}$ such that $|f(z)|$ is a perfect square is $2^m$, where $m=\#$ of distinct prime divisors of $\Delta_f$.

The question can be equated to the counting of solutions to the following diophantine equation: $$4a^4x^2\pm y^2=\Delta_f,$$ where $\Delta_f=b^2-4a^2c$ and plus\minus relate to considering when $f(x)=-\square$ or $\square$. Thus we have a Ramanujan–Nagell type equation.

In the special case when $a=1$ more can be said. In particular, for counting negatives squares, there are known formulas for the $r_2(\Delta_f)$.