Exercise 2.6.5 (Solve $x^3 + 10x^2 +x + 3 \equiv 0 \pmod{3^3}$)

Question

Solve $x^3 + 10x^2 +x + 3 \equiv 0 \pmod{3^3}$

Richard Ganaye · Accepted Answer

\begin{proof} The solutions of $x^3 + 10x^2 +x + 3 \equiv 0 \pmod{3}$ are $0,1$ modulo $3$. Moreover $f'(x) = 3x^2 + 20x + 1$, then $f'(0) =\equiv 1 \pmod 3$ and $f'(1) \equiv 0 \pmod 3$, so $0$ is a nonsingular root, but $1$ is singular.

The nonsingular solution $a_1 = 0$ lifts to a unique solution $a_2 $ modulo $9$, and an unique solution $a_3$ modulo $27$, where 
\begin{align*}
a_2 &= a_1  - f(0) \overline{f'(0)} = -3 \equiv 6 \pmod 9,\
a_3 &=a_2 -f(a_2) \overline{f'(0)} = 6 - 585 \equiv 15 \pmod {27}.
\end{align*}
The singular solution $b_1 = 1$ satisfies, using (2.4),
$$ f(1 + t p) \equiv f(1) = 15 \equiv 6 \pmod 9.$$
Since $f(1) = 15 
ot \equiv 0 \pmod 9$, $1$ does not lift to any solution modulo $9$, a fortiori modulo $27$.

Therefore 
$$x^3 + 10x^2 +x + 3 \equiv 0 \pmod{3^3} \iff x \equiv 15 \pmod{3^3}.$$
\end{proof}

Check with Sage:
\begin{verbatim}
Z27 =  IntegerModRing(27)
R.<x> = PolynomialRing(Z27)
f(x) = x^3 + 10*x^2 + x + 3
[a for a in Z27 if f(a) == 0]
          [15]
\end{verbatim}

Exercise 2.6.5 (Solve $x^3 + 10x^2 +x + 3 \equiv 0 \pmod{3^3}$)

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Exercise 2.6.5 (Solve $x^3 + 10x^2 +x + 3 \equiv 0 \pmod{3^3}$)

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