# How many solutions to $x^2-x+5\equiv 0\pmod{p^2}$

Let $p$ be a prime. How many solutions modulo $p$ has the equation $x^2-x+5\equiv 0\pmod{p^2}$.

My thoughts: I first consider equation modulo $p$. I've found:

• no solutions for $p=2$

• one solution for $p=19$

• two solutions for $p\in A:=\{1,4,5,6,7,9,11,16,17\}\pmod{19}$

• no solutions otherwise

Now, consider the equation modulo $p^2$. Obviously, if $x$ is a solution of $x^2-x+5\pmod{p^2}$ then it is a solution of $x^2-x+5=0\pmod{p}$. Hence, there are no solutions when $p=2$, and when $p\notin A$. Let $p$ be odd, then the equation can be written as $(2x-1)^2=-19\pmod{p^2}$. This equality proves immediately that there are no solutions when $p=19$. Suppose $p\in A$. Then there are two solutions of $(2x-1)^2=-19\pmod{p}$. Let $a$ be such a solution. Then $x=a+yp$, some $y\in\mathbb{Z}$, and substituting this equality we have $$(2a-1)y=-\frac{a^2-a+5}{p}\pmod{p}$$

This is an equation for $y$, which has a unique solution since $2a-1\neq 0\pmod{p}$. I would like to deduce from this that each solution of the equation $\pmod{p}$ lifts to a unique solution of the equation $\pmod{p^2}$, so that for $p\in A$ they both have 2 solutions. How can I deduce this?

• It seems to me that you have already deduced what you say you want to deduce. For each solution modulo $p$, you have found a unique solution modulo $p^2$. What remains to be deduced? – Gerry Myerson Sep 25 '13 at 13:18
• @GerryMyerson I mean, let $a$ be a solution of equation mod $p$, so that $p$ divides $(2a-1)^2+19$. Now let $b:=a +yp$, with $y$ uniquely determined as in my question. Why is it obviuos that $p^2$ divides $(2b-1)^2+19$?? – Danae Kissinger Sep 25 '13 at 13:32
• You chose $y$ so that you'd have $x^2-x+5\equiv0\pmod{p^2}$, didn't you? – Gerry Myerson Sep 25 '13 at 13:35
• @GerryMyerson Ok, given $a$ solution modulo $p$, I've found one and only one solution of the form $a+yp$ of the equation modulo $p^2$. Why is it unique modulo $p^2$? Is it because $p$ divides $y$? – Danae Kissinger Sep 25 '13 at 13:52