Exercise 2.6.2 (Solve $x^5 + x^4 + 1 \equiv 0 \pmod{3^4}$)

Question

Solve $x^5 + x^4 + 1 \equiv 0 \pmod{3^4}$

Richard Ganaye · Accepted Answer

\begin{proof} 
We first solve $f(x) = x^5 + x^4 + 1 \equiv 0 \pmod{3}$. Since $f(0) = f(2) \equiv 1 \pmod{3}$, the only solution is $x_1 = 1$.

Therefore every solution $x_1$ of $f(x) = x^5 + x^4 + 1 \equiv 0 \pmod{9}$ is of the form
$$ x_1 = 1 + 3t.$$
Here $f'(x) = 5x^4 + 4 x^3$, $f'(1) \equiv 0 \pmod 3$, so $1$ is a singular solution.

In this case, the Taylor's expansion (2.4) gives
$$f(1+ 3t) \equiv f(1) \pmod{3^2}.$$
But $f(1) = 3 
ot \equiv 0 \pmod{9}$. This shows that $x^5 + x^4 + 1 \equiv 0 \pmod{3^2}$ has no solution. A fortiori, $x^5 + x^4 + 1 \equiv 0 \pmod{3^4}$ has no solution.
\end{proof}

Check (with Sage):
\begin{verbatim}
Z81 = IntegerModRing(81)
R.<x> = PolynomialRing(Z81)
f = x^5 + x^4 + 1
[a for a in Z81 if f(a) == 0]
         []
\end{verbatim}

Exercise 2.6.2 (Solve $x^5 + x^4 + 1 \equiv 0 \pmod{3^4}$)

Answers

Comments

Exercise 2.6.2 (Solve $x^5 + x^4 + 1 \equiv 0 \pmod{3^4}$)

Answers

Comments

Add answer