Exercise 2.3.14 (Solve $x^3 + 2x - 3 \equiv 0 \pmod{45}$)

Proof. In $\mathbb{Z}[x]$,

a)

This shows that $x^3+2x-3\equiv 0(\mathit{mod}9)$ is equivalent to $$\begin{array}{lll}\hfill & 9\mid x-1& \hfill \text{(1)}\\ \hfill \text{or }& 9\mid x^2+x+3& \hfill \text{(2)}\\ \hfill \text{or }& (3\mid x-1,3\mid x^2+x+3)& \hfill \text{(3)}\end{array}$$

(3) has no solution, because $3\mid x-1$ and $3\mid x^2+x+3$ imply $x\equiv 1(\mathit{mod}3),0\equiv x^2+x+3\equiv 2(mod3)$, which gives the contradiction $0\equiv 2(\mathit{mod}3)$.

Now we solve (2).

$$\begin{array}{llll}\hfill x^2+x+3\equiv 0(\mathit{mod}9)& ⟺{(x+5)}^2-22\equiv 0(\mathit{mod}9)& \hfill & \\ \hfill & ⟺{(x+5)}^2-4\equiv 0(\mathit{mod}9)& \hfill & \\ \hfill & ⟺(x-6)(x-2)\equiv 0(\mathit{mod}9).& \hfill & \end{array}$$

It is impossible that $3\mid x-6$ and $3\mid x-2$ simultaneously, otherwise $0\equiv x\equiv 2(\mathit{mod}3)$. Therefore the last equation is equivalent to $x\equiv 2$ or $x\equiv 6(\mathit{mod}9)$. So we obtain

$$x^3+2x-3\equiv 0(\mathit{mod}9)⟺x\equiv 1,2,6(mod9).$$

b)

Since $5$ is prime, $$\begin{array}{llll}\hfill x^3+2x-3\equiv 0(\mathit{mod}5)& ⟺(x-1)(x^2+x+3)\equiv 0(\mathit{mod}5)& \hfill & \\ \hfill & ⟺x\equiv 1(\mathit{mod}5)\text{ or }{x}^2+x+3\equiv 0(mod5)& \hfill & \\ \hfill & ⟺x\equiv 1(\mathit{mod}5)\text{ or }(x-3)(x-1)\equiv 0(mod5)& \hfill & \\ \hfill & ⟺x\equiv 1(\mathit{mod}5)\text{ or }x\equiv 3(mod5).& \hfill & \end{array}$$

(In ${𝔽}_5[x]$, $x^3+2x-3={(x-1)}^2(x-3)$.)

c)

Combining part (a) and (b), we obtain $$\begin{array}{llll}\hfill & x^3+2x-3\equiv 0(\mathit{mod}45)& \hfill & \\ \hfill & ⟺x^3+2x-3\equiv 0(\mathit{mod}5)\text{ and }{x}^3+2x-3\equiv 0(mod9)& \hfill & \\ \hfill & ⟺x\equiv 1,3(\mathit{mod}5)\text{ and }x\equiv 1,2,6(mod5)& \hfill & \\ \hfill & ⟺\{\begin{array}{c}x\equiv 1(\mathit{mod}5)\hfill \\ x\equiv 1(\mathit{mod}9)\hfill \end{array}\text{ or }\{\begin{array}{c}x\equiv 1(\mathit{mod}5)\hfill \\ x\equiv 2(\mathit{mod}9)\hfill \end{array}\text{ or }\{\begin{array}{c}x\equiv 1(\mathit{mod}5)\hfill \\ x\equiv 6(\mathit{mod}9)\hfill \end{array}& \hfill & \\ \hfill & \text{ or }\{\begin{array}{c}x\equiv 3(\mathit{mod}5)\hfill \\ x\equiv 1(\mathit{mod}9)\hfill \end{array}\text{ or }\{\begin{array}{c}x\equiv 3(\mathit{mod}5)\hfill \\ x\equiv 2(\mathit{mod}9)\hfill \end{array}\text{ or }\{\begin{array}{c}x\equiv 3(\mathit{mod}5)\hfill \\ x\equiv 6(\mathit{mod}9)\hfill \end{array}& \hfill & \\ \hfill & ⟺x\equiv 1,11,6,28,38,33(\mathit{mod}45).& \hfill & \end{array}$$

The solutions of $x^3+2x-3\equiv 0(\mathit{mod}45)$ are $x\equiv 1,6,11,28,33,38(\mathit{mod}45)$.