# Poincare series

I understand the concepts used in then Poincare series, but I don't know how to compute the Poincare serie of an specific polinomyal, for example $x^2-a$, what must I do?

-
It's my understanding that the Poincare series of a polynomial $f$ is $\sum a_nx^n$, where $a_n$ is the number of solutions of $f(x)\equiv0\pmod{p^n}$ --- but you have to be given a prime $p$ to make this definition go. Anyway, let's start with this question: given a prime $p$, what can you say about the number of solutions of $x^2-a\equiv0\pmod p$? –  Gerry Myerson Nov 15 '12 at 4:42

Let's work out your example for $f(x)=x^2-a \in \mathbb{Z}[x]$. For a prime $p>2$, suppose first that $x^2 \equiv a \mod p$ has no solutions (i.e. $a$ is not a residue mod $p$). Then $x^2 \equiv a \mod p^n$ has no solutions for any $n \geq 1$, as these would project down to roots of $x^2-a \mod p$. In this case, we find that the Poincare series is $0$.
If instead $x^2 \equiv a \mod p$ admits a root $r \not\equiv 0 \mod p$, the fact that $f'(r) =2r \not\equiv 0 \mod p$ implies by Hensel's lemma that there is a unique solution $s$ to $x^2 \equiv a \mod p^2$ such that $s \equiv r \mod p$. Thus our two solutions mod $p$ lift to two solutions mod $p^2$, and so on (again by Hensel's lemma). So we always obtain two solutions, and our series is $$\sum_{n=1}^\infty 2 x^n = \frac{2x}{1-x}.$$
Else $a\equiv 0 \mod p$, a situation that requires more work. Write $a=p^k m$ with $(p,m)=1$. Then $x^2 \equiv p^km \mod p^n$ (with $n$ sufficiently large) forces $x$ to be divisible by $p$, with multiplicity $\lceil k/2 \rceil$. Dividing through yields $$(x/p^{\lceil k/2 \rceil})^2 \equiv m \mod p^{n-\lceil k/2\rceil},$$ (so in fact $k$ must be even to obtain solutions mod $p^n$ with $n$ large). For $y:=x/p^{\lceil k/2 \rceil}$, we see that this problem can be solved with one of the cases above.
Note that some of the above work falls apart for $p=2$.