Exercise 2.6.7 (Solve $x^3 + x^2 - 5 \equiv 0 \pmod{7^3}$)

Question

Solve $x^3 + x^2 - 5 \equiv 0 \pmod{7^3}$.

Richard Ganaye · Accepted Answer

\begin{proof} 
The equation $x^3 + x^2 - 5 \equiv 0 \pmod{7}$ has a unique solution $2$. Moreover $f'(x) = 3x^2 + 2x$, so $f'(2) = 16 \equiv 2 
ot \equiv 0 \pmod{7}$, so $2$ is a nonsingular root. Therefore $a = a_1 = 2$ lifts to a unique solution $a_2$ modulo $7^2$, and a unique solution $a_3$ modulo $7^3$, where
\begin{align*}
a_2 &= a_1 - f(a_1) \overline{f'(a)} = 2 - 7 	imes 4 = -26 \equiv 23 \pmod {7^2},\
a_3 &= a_2 - f(a_2)\overline{f'(a)} = 23 - 12691 	imes 4 =  23 - 7^3\cdot 37 \cdot 4 \equiv 23 \pmod{7^3}.
\end{align*}
This gives
$$x^3 + x^2 - 5 \equiv 0 \pmod{7^3} \iff x \equiv 23 \pmod{7^3}.$$
\end{proof}

Exercise 2.6.7 (Solve $x^3 + x^2 - 5 \equiv 0 \pmod{7^3}$)

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Exercise 2.6.7 (Solve $x^3 + x^2 - 5 \equiv 0 \pmod{7^3}$)

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