Number of Roots of a Quadratic Modulo Prime Powers

I am an undergraduate and have been working on some research that I plan to publish. Part of the proof for one of my theorems relies on proving that an equation of the form $ax^2 + bx + c \equiv 0 \pmod{p^y}$ ($p$ an odd prime and $y > 1$ if that ends up mattering) is not "trivial" (by which I mean not every $x$ is a solution) for $x < p$.

My current understanding is that a polynomial of degree $d$ over a finite field either is identically true in the field or has not more than $d$ roots in the field (please correct me if this is wrong). Without getting into what $a,b,c$ are in my proofs (it really doesn't matter for this question), I have proven that $a$ is relatively prime to $p$ (and thus relatively prime to the modulus).

Shouldn't this be sufficient to prove that this polynomial isn't "trivial"? And one step further, the data I've looked at suggests that this polynomial should always have either 0 or two roots - is this sufficient to prove that?

• Are you working $\mod p^y$ or over a finite field of order $p^y$? Because if you just work $\mod p^y$ then a quadratic equation might have more than two roots. For example, $x^2 - 1 \equiv 0 \, (\mod 8)$ is satisfied by $x = 1,3,5,7$. Also, if you want to show the equation is "non-trivial", can't you start with showing that $c \neq 0$ ($\mod p^y$)? Because then $x = 0$ won't be a solution and if $c = 0 (\mod p^y)$ you can just through it away from the equation. Aug 13, 2017 at 16:42
• "My current understanding is that a polynomial of degree $d$ over a finite field either is identically true in the field or has not more than $d$ roots in the field (please correct me if this is wrong)." You seem to confuse a finite field with integers modulo a prime power. Which one do you want to ask about? Aug 13, 2017 at 16:43
• Consider $x^2 \equiv 0 \bmod 9$ then. It has three roots, $x=0,3,6$. Aug 13, 2017 at 16:51
• You probably want to familiarize yourself with Hensel lifting. If $x=n$ is a zero of a polynomial $f(x)$ modulo $p$, and $f'(n)$ is not divisible by $p$, then there is a unique solution $x=n'$ of $f(x)\equiv\pmod{p^k}$ such that $n'\equiv n \pmod p$. Modulo $p$ you can take advantage of the fact that $\Bbb{Z}/p\Bbb{Z}$ is a field. Aug 13, 2017 at 16:55
• @hardmath I suppose it would help here to give some context on the actual problem. This polynomial is being used to, put simply, extend the integers modulo $p^y$ to the integers modulo $p^{y+1}$. So the purpose of the variable $x$ is to take each integer modulo $p^y$ to the $p$ integers equivalent to it modulo $p^{y+1}$. And for more complicated reasons, I don't really care about the case where $x = 0$, so only roots relatively prime to $p$ matter. Does this change things? Aug 13, 2017 at 17:00