How to solve the congruence $ x^4 + x + 3 = 0 \pmod{3^3}$?

Question

$$x^4 + x + 3 = 0 \pmod{3^3}$$ I'm not sure how to this, I've tried many times but it never works for me :/ so, I hope someone will help me

Answer 1

As the modulus is small, it isn't too hard to just check all the possible values of $x$. For $x = 0, 1, 2, 3$, the calculations are easy enough:

\begin{align*} x = 0 &\implies x^4 + x + 3 = 3 \not\equiv 0 \bmod 9\\ x = 1 &\implies x^4 + x + 3 = 5 \not\equiv 0 \bmod 9\\ x = 2 &\implies x^4 + x + 3 = 21 \not\equiv 0 \bmod 9\\ x = 3 &\implies x^4 + x + 3 = 87 \not\equiv 0 \bmod 9. \end{align*}

For $x \geq 4$, the presence of $x^4$ makes calculations slightly less trivial (not only calculating the value of $x^4 + x + 3$, but also determining whether the result is divisible by $9$). One way to proceed is to first work out $x^2 \bmod 9$ then square the result to obtain $x^4 \bmod 9$.

\begin{align*} x = 4 &\implies x^2 = 16 \equiv 7 \bmod 9 \implies x^4 \equiv 4 \bmod 9\\ x = 5 &\implies x^2 = 25 \equiv 2 \bmod 9 \implies x^4 \equiv 4 \bmod 9\\ x = 6 &\implies x^2 = 36 \equiv 0 \bmod 9 \implies x^4 \equiv 0 \bmod 9\\ x = 7 &\implies x^2 = 49 \equiv 4 \bmod 9 \implies x^4 \equiv 7 \bmod 9\\ x = 8 &\implies x^2 = 64 \equiv 1 \bmod 9 \implies x^4 \equiv 1 \bmod 9. \end{align*}

With these results at hand, the calculations becomes easier.

\begin{align*} x = 4 &\implies x^4 + x + 3 \equiv 4 + 4 + 3 \bmod 9 \not\equiv 0 \bmod 9\\ x = 5 &\implies x^4 + x + 3 \equiv 4 + 5 + 3 \bmod 9 \not\equiv 0 \bmod 9\\ x = 6 &\implies x^4 + x + 3 \equiv 0 + 6 + 3 \bmod 9 \equiv 0 \bmod 9\ \ \checkmark\\ x = 7 &\implies x^4 + x + 3 \equiv 7 + 7 + 3 \bmod 9 \not\equiv 0 \bmod 9\\ x = 8 &\implies x^4 + x + 3 \equiv 1 + 8 + 3 \bmod 9 \not\equiv 0 \bmod 9. \end{align*}

So the only solution is $x = 6$.