已知 $a,b,c$ 为整数，$24\mid a^3+b^3+c^3$，求证：$120\mid a^5+b^5+c^5+4(a+b+c)$．
解析 令 $x_1=a$，$x_2=b$，$x_3=c$，$A=\displaystyle\sum_{i=1}^{3}x_i^3$，$B=\displaystyle\sum_{i=1}^{3}\left(x_i^5+4x_i\right)$，则 \[\begin{split} B-5A &=\displaystyle\sum_{i=1}^{3}\left(x_i^5+4x_i-5x_i^3\right)\\ &=\displaystyle\sum_{i=1}^{3}\left(x_i^3-x_i\right)\left(x_i^2-4\right)\\ &=\displaystyle\sum_{i=1}^{3}\big[x_i\left(x_i+1\right)\left(x_i-1\right)\left(x_i+2\right)\left(x_i-2\right)\big]. \end{split}\] 由于 $5$ 个连续整数的乘积能被 $5!=120$ 整除（考虑组合数 $\mathrm{C}_{n}^{5}$），所以\[120\mid B-5A,\]而 $120\mid 5A$，故\[120\mid B.\]