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Prove that for any integer n, 5 divides n^5-n.
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= = = = .
If n is a multiple of 5, the statement is true.
If n gives the remainder 1 when divided by 5, then the factor (n-1) is a multiple of 5, and the statement is true.
If n gives the remainder 4 when divided by 5, then the factor (n+1) is a multiple of 5, and the statement is true.
If n gives the remainder 2 when divided by 5, then the factor (n^2+1) is a multiple of 5.
Indeed, the remainder of division by 5 is = 5 (equivalent to 0) in this case, and the statement is true.
If n gives the remainder 3 when divided by 5, then the factor (n^2+1) is a multiple of 5.
Indeed, the remainder of division by 5 is = 10 (equivalent to 0) in this case, and the statement is true.
Thus the statement is true in all cases.
QED. Proved and solved.