<s>[INST] How to solve the congruence $ x^4 + x + 3 = 0 \pmod{3^3}$? $$x^4 + x + 3 = 0 \pmod{3^3}$$
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
 [/INST] 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 $x = 6$.
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