Find all integers $x$ such that $7 x^{5}+x^{3}-x$ +2 is divisible by 120.

The original text has been translated into English while preserving the original line breaks and format.
Solution: Let $f(x)=7 x^{5}+x^{3}-x+2$.
(1) $120=2^{3} \times 3 \times 5$, where $2^{3}, 3, 5$ are pairwise coprime.
(2) $3|f(x) \Leftrightarrow 3|\left(x^{5}+x^{3}-x-1\right)$.

Obviously, $3 \mid x(x-1)(x+1)=x^{3}-x$, so
$$
\begin{array}{l}
3|f(x) \Leftrightarrow 3|\left(x^{5}-1\right) \\
\Leftrightarrow 3 \mid\left[\left(x^{2}+1\right)\left(x^{3}-x\right)+x-1\right] \\
\Leftrightarrow 3 \mid(x-1) .
\end{array}
$$
(3) $5|f(x) \Leftrightarrow 5|\left(2 x^{5}+x^{3}-x+2\right)$.

Obviously, $5 \mid x(x-1)(x+1)(x-2)(x+2)$
$$
\Leftrightarrow 5 \mid\left(x^{5}-5 x^{3}+4 x\right) \Leftrightarrow 5 \mid\left(x^{5}-x\right) \text {. }
$$

Thus, $5|f(x) \Leftrightarrow 5|\left(x^{3}+x+2\right)$
$$
\Leftrightarrow 5 \mid(x+1)\left(x^{2}-x+2\right) \text {. }
$$

Substituting $x=0,1,2,3,4$ into $x^{2}-x+2$, we find that $5 \mid\left(x^{2}-x+2\right)$.
Thus, $5 \mid f(x) \Leftrightarrow 5 \mid(x+1)$.
(4) $2^{3}\left|f(x) \Leftrightarrow 2^{3}\right|\left(x^{5}-x^{3}+x-2\right)$.

Obviously, $2 \mid x$, so
$$
2^{3} \mid f(x) \Leftrightarrow 2^{3} \mid (x-2) \text {. }
$$
(5) From equation (1), we get $15 \mid (5 x-5)$.

From equation (2), we get $15 \mid (3 x+3)$.
Thus, $15 \mid [2(5 x-5)-3(3 x+3)] \Leftrightarrow 5 \mid (x-19)$.
From equation (3), we get $120 \mid (15 x-30)$.
From equation (4), we get $120 \mid (8 x-152)$.
Thus, $120 \mid [2(8 x-152)-(15 x-30)]$
$$
\Leftrightarrow 120 \mid (x-274) \Leftrightarrow 120 \mid (x-34) \text {. }
$$

Therefore, $x=120 k+34(k \in \mathbb{Z})$.
(Wu Weizhao, School of Mathematics and Information Science, Guangzhou University, 510006)