I have a polynomial $f(x)=a^2x^2+bx+c\in\mathbb Z[x]$ with $f(x)$ not a constant times a square and $abc\neq0$ and I want to know how many $x$ between $-a$ and $a$ the polynomial is a perfect square. Since leading coefficient is a perfect square by this reference the polynomial only has finitely many squares.

  1. How many squares can be achieved by $f(x)$ with $x\in(-a,a)\cap\mathbb Z$ if $|a|^6,|b|^4,|c|^3$ are roughly of similar size? Gap between squares is squareroot in size and $(x-y)|(f(x)-f(y))$ makes me think there is at most only $1$ square with $x\in(-a,a)\cap\mathbb Z$.

  2. Is there a way to test in polynomial time the parity of number of square $f(x)$ with $x\in(-t,t)\cap\mathbb Z$ at any given $t<a$ or find at least one such square?

Is there any good reference on the topic? It suffices to compute $\sum_{x=-t}^t\omega(f(x))\bmod 2$ where $\omega(f(x))$ is number of divisors of $f(x)$.


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 $r_2(\Delta_f)$.

  • $\begingroup$ The perfect square solutions correspond to the integer solutions to the equation $\Delta_f=x^2+y^2$, i.e. $2r'_2(\Delta_f)$. Given that we require a squarefree discriminant, we cannot have any prime divisors that are congruent to $3 \pmod{4}$. $\endgroup$ – pavl0 Aug 15 '18 at 8:19
  • $\begingroup$ 'The perfect square solutions correspond to the integer solutions to the equation $\Delta_f=x^2+y^2\dots$' why? I do not see this. $\endgroup$ – 1.. Aug 15 '18 at 9:17
  • $\begingroup$ If $f(z)=-\square$ (note that f(z) is always negative for $z\in (\alpha_1,\alpha_2)$), then letting $z'=z+\frac{b}{2}$, we get $f(z)=z'^2-\frac{\Delta_f}{4}$, thus $\frac{\Delta_f}{4}=\square+z'^2.$ $\endgroup$ – pavl0 Aug 15 '18 at 12:14
  • $\begingroup$ $f(z)$ is $f(x)$ evaluated at $z$. For example, $f(x)=x^2-3x+1$, then $f(1)=-1$. (Also, I assumed that $f(x)$ is irreducible.) $\endgroup$ – pavl0 Aug 15 '18 at 16:45
  • 1
    $\begingroup$ I do not see the connection en.wikipedia.org/wiki/Ramanujan–Nagell_equation#Equations_of_Ramanujan–Nagell_type or the reduction. $\endgroup$ – 1.. Aug 16 '18 at 8:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.