Exercise 2.9.3* (Number of solutions modulo $p^2$)

Note first that if $a{x}^2+\mathit{bx}+c\equiv 0\mathit{mod}{p}^2$, then $a{x}^2+\mathit{bx}+c\equiv 0\mathit{mod}p$

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Assume first that $p\nmid a$, and $p\nmid D=b^2-4\mathit{ac}$. Then the equation $f(x)\equiv 0(\mathit{mod}p)$ has two distinct solutions $x_0,x_0^{\prime }$ modulo $p$ (see Exercise 2.9.2). If $x_1$ is a solution of $f(x)\equiv 0(\mathit{mod}{p}^2)$, then $f(x_1)\equiv 0(\mathit{mod}{p}^2)$, a fortiori $f(x_1)\equiv 0(\mathit{mod}p)$, thus $x_1\equiv x_0(\mathit{mod}p)$ or $x_1\equiv x_0^{\prime }(\mathit{mod}p)$.

If $x_1\equiv x_0(\mathit{mod}p)$, then $x_1=x_0+\mathit{kp},k\in \mathbb{Z}$, where $a{x}_0^2+b{x}_0+c=\mathit{\lambda p},\lambda \in \mathbb{Z}$. Since

$$\begin{array}{llll}\hfill f(x_1)& =a{(x_0+\mathit{kp})}^2+b(x_0+\mathit{kp})+c& \hfill & \\ \hfill & =(a{x}_0^2+b{x}_0+c)+(2a{x}_0+b)\mathit{kp}& \hfill & \\ \hfill & =\mathit{\lambda p}+(2a{x}_0+b)\mathit{kp},& \hfill & \end{array}$$

we obtain

$$\begin{array}{llll}\hfill f(x_1)\equiv 0(\mathit{mod}{p}^2)& ⟺\mathit{\lambda p}+(2a{x}_0+b)\mathit{kp}\equiv 0(\mathit{mod}{p}^2)& \hfill & \\ \hfill & ⟺\lambda +f^{\prime }(x_0)k\equiv 0(\mathit{mod}p).& \hfill & \end{array}$$

By Exercise 2.9.2, in the case $p\nmid a,p\nmid D$, we know that $p\nmid f^{\prime }(x_0)$. Therefore the equation $\lambda +f^{\prime }(x_0)k\equiv 0(\mathit{mod}p)$ has a unique solution $k$.

We can conclude that the equation $f(x)\equiv 0(\mathit{mod}{p}^2)$ has two solutions modulo $p^2$, which are $x_0+\mathit{kp},x_0^{\prime }+k^{\prime }p$, where $k\equiv -\lambda f^{\prime }{(x_0)}^{-1}(\mathit{mod}p),k^{\prime }\equiv -{\lambda }^{\prime }{f}^{\prime }{(x_0^{\prime })}^{-1}(modp)$, and $\lambda =f(x_0)∕p,{\lambda }^{\prime }=f(x_0^{\prime })∕p$.

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Assume now that $p\nmid a$, but $p\mid D=b^2-4\mathit{ac}$. Then the equation $f(x)\equiv 0(\mathit{mod}p)$ has a unique solution $x_0=2^{-1}{a}^{-1}b$ mod $p$, where $f^{\prime }(x_0)\equiv 0(\mathit{mod}p)$ (see Exercise 2.9.2).

A solution $x_1$ of $f(x)\equiv 0(\mathit{mod}{p}^2)$ is of the form $x_1=x_0+\mathit{kp},k\in \mathbb{Z}$. Since $f(x_0)=a{x}_0^2+b{x}_0+c=\mathit{\lambda p}$ for some $\lambda \in \mathbb{Z}$, and $f^{\prime }(x_0)=0$, we obtain

$$\begin{array}{llll}\hfill f(x_1)\equiv 0(\mathit{mod}{p}^2)& ⟺\lambda \equiv 0(\mathit{mod}p)& \hfill & \\ \hfill & ⟺p^2\mid f(x_0).& \hfill & \end{array}$$

If $p^2\nmid f(x_0)$, there is no solution, and if $p^2\mid f(x_0)$ any value of $k$ gives a solution, so that there are $p$ solutions $x_0,x_0+p,\cdots ,x_0+(p-1)p$ distinct modulo $p^2$.

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if $p\mid a,p\nmid b$, the equation $f(x)=\mathit{bx}+c\equiv 0(\mathit{mod}p)$ has a unique solution $x_0\equiv -c{b}^{-1}(\mathit{mod}p)$. if $x_1=x_0+\mathit{kp}$ is a solution of $f(x)\equiv 0(\mathit{mod}{p}^2)$, then $b{x}_0+c=\mathit{\lambda p}$ for some integer $\lambda$. $$\begin{array}{llll}\hfill f(x_1)\equiv 0(\mathit{mod}{p}^2)& ⟺b(x_0+\mathit{kp})+c\equiv 0(\mathit{mod}{p}^2)& \hfill & \\ \hfill & ⟺\mathit{\lambda p}+\mathit{bkp}\equiv 0(\mathit{mod}{p}^2)& \hfill & \\ \hfill & ⟺\lambda +\mathit{bk}\equiv 0(\mathit{mod}p).& \hfill & \end{array}$$

Since $p\nmid b$, this equation has exactly one solution modulo $p$, which give a unique solution $x_0+\mathit{kp}$ of $f(x)\equiv 0(\mathit{mod}{p}^2)$.

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If $p\mid a,p\mid b$, then $p\nmid c$, and the equation $f(x)=c\equiv 0\mathit{mod}{p}^2$ has no solution.

To conclude, the number $N$ of solutions modulo $p^2$ of $a{x}^2+\mathit{bx}+c\equiv 0\mathit{mod}{p}^2$ is given by

where $x_0=2^{-1}{a}^{-1}b$.