Exercise 2.6.1 (Solve $x^2+x+7 \equiv 0 $ modulo $3^3$ and modulo $3^4$)

Proof.

a)

Since $4\wedge 27=1$, $$\begin{array}{llll}\hfill x^2+x+7\equiv 0(\mathit{mod}27)& ⟺4x^2+4x+28\equiv 0(\mathit{mod}27)& \hfill & \\ \hfill & ⟺{(2x+1)}^2+27\equiv 0(\mathit{mod}27)& \hfill & \\ \hfill & ⟺{(2x+1)}^2\equiv 0(\mathit{mod}27).& \hfill & \end{array}$$

Write $a=2x+1$. If $a^2\equiv 0(\mathit{mod}{3}^3)$, then $2{\nu }_3(a)\ge 3$, so ${\nu }_3(a)\ge 3∕2$. Since ${\nu }_3(a)$ is an integer, ${\nu }_3(a)\ge 2$, thus $a\equiv 0(\mathit{mod}{3}^2)$. Conversely, if $a\equiv 0(\mathit{mod}{3}^2)$, then $a^2\equiv 0(\mathit{mod}{3}^4)$, a fortiori $a^2\equiv 0(\mathit{mod}{3}^3)$. This shows that

$$\begin{array}{lll}\hfill a^2\equiv 0(\mathit{mod}{3}^3)⟺a\equiv 0(mod{3}^2).& & \hfill \text{(1)}\end{array}$$

Therefore

$$\begin{array}{llll}\hfill x^2+x+7\equiv 0(\mathit{mod}27)& ⟺{(2x+1)}^2\equiv 0(\mathit{mod}27)& \hfill & \\ \hfill & ⟺2x+1\equiv 0(\mathit{mod}9)& \hfill & \\ \hfill & ⟺x\equiv 4(\mathit{mod}9)& \hfill & \\ \hfill & ⟺x\equiv 4,13,22(\mathit{mod}27).& \hfill & \end{array}$$

We find the same solutions than in Example 12:

$$x^2+x+7\equiv 0(\mathit{mod}27)⟺x\equiv 4,13,22(mod27).$$

b)

Similarly, $$\begin{array}{llll}\hfill x^2+x+7\equiv 0(\mathit{mod}81)& ⟺4x^2+4x+28\equiv 0(\mathit{mod}81)& \hfill & \\ \hfill & ⟺{(2x+1)}^2+27\equiv 0(\mathit{mod}81).& \hfill & \\ \hfill & & \hfill \end{array}$$

Then ${(2x+1)}^2\equiv 0(\mathit{mod}27)$. By (1), $2x+1\equiv 0(\mathit{mod}{3}^2)$. If we divide all terms by $27=3^3$, we obtain

$$\begin{array}{llll}\hfill x^2+x+7\equiv 0(\mathit{mod}81)& ⟺3{\left(\frac{2x+1}{3^2}\right)}^2+1\equiv 0(\mathit{mod}3)& \hfill & \\ \hfill & ⟺1\equiv 0(\mathit{mod}3).& \hfill & \end{array}$$

This contradiction shows that the congruence $x^2+x+7\equiv 0(\mathit{mod}81)$ has no solution.