Ta có: \(x^5-x=x\left(x^4-1\right)=x\left(x^2-1\right)\left(x^2+1\right)=x\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\)
\(=x\left(x-1\right)\left(x+1\right)\left(x^2-4\right)+5x\left(x-1\right)\left(x+1\right)\)
\(=\left(x-2\right)\left(x-1\right)x\left(x+1\right)\left(x+2\right)+5x\left(x-1\right)\left(x+1\right)\)
Vì \(\left(x-2\right)\left(x-1\right)x\left(x+1\right)\left(x+2\right)\)là tích 5 số nguyên liên tiếp suy ra chia hết cho 5
\(5x\left(x-1\right)\left(x+1\right)\)chia hết cho 5
\(\Rightarrow\left(x-2\right)\left(x-1\right)x\left(x+1\right)\left(x+2\right)+5x\left(x-1\right)\left(x+1\right)\)chia hết cho 5 hay \(x^5-x\)chia hết cho 5(điều phải chứng minh)
