Let $P(x)$ be a polynomial with integer coefficients, and $p$ be a large prime. I want to find the number of solutions to the congruence equation $$P(x)\equiv 0(\mathrm{mod}\ p^\alpha).$$
In my practice the degree of $P(x)$ is $4$. We can write $$ P(x) = \sum_{j=0}^4 a_j x^j.$$ I wonder that whether the number of solutions is at most $4$ when $\mathrm{gcd}(a_0,a_1,a_2,a_3,a_4)=1$?
If not, is there a finite upper bound on the number of solutions (not depend on $p$)?
Thanks in advance for your help!
COMMENT.- (Just for help you a little).
$(1)\space\space P(x)\equiv0\pmod{p^{\alpha}}\Rightarrow P(x)\equiv0\pmod p$. If $P(x)$ has degree $4$ and has more than $4$ solutions, then all the coefficients of $P(x)$ are multiples of $p$ (i.e. are $0$ modulo $p$)
$(2)$ Let $P(x_1)\equiv0\pmod p$ and put $x=x_1+pt_1$ where $t_1\in\mathbb Z$ then look at the equation $P(x_1+pt_1)\equiv0\pmod{p^2}$. Using Taylor's series for your polynomial $P(x)$, you do have $$P(x_1)+pt_1P'(x_1)\equiv0\pmod{p^2}$$ here you could have trouble if $P'(x_1)\equiv0\pmod{p}$. If $P'(x_1)\not\equiv0\pmod{p}$, you have a solution $t_1=t'_1\pmod p$ so $x$ becomes $$x=x_1+pt'_1+p^2t_2=x_2+p^2t_2$$ and you have the equation $$P(x)\equiv0\pmod{p^3}$$ At this point $P'(x_2)\not\equiv0\pmod p$ is not divisible by $p$ (Why?) so this last equation have a unique solution $t_2$ with $t_2=t'_2\pmod p$. Now you do have $$x=x_2+p^2t'_2+p^3t_3=x_3+p^3t_3$$ and so on.
►If $P'(x_1)\not\equiv0\pmod{p}$ then $x_1$ leads to a solution of $P(x)\equiv0\pmod{p^\alpha}$.
Try with the equation $$x^4+5x+7\equiv0\pmod{27}\text { whose solution is } x\equiv{23}\pmod{27}$$