Problem
NT-B2-M04-P006 The Form \(a^p-a\)
#6
★★★☆☆ Level 3 of 5
Prove that for every prime \(p\) and every integer \(a\), \(p\mid a^p-a\).
Consider the cases \(p\mid a\) and \(p\nmid a\).
If \(p\mid a\), then \(a^p-a\) is divisible by \(p\). If \(p\nmid a\), Fermat gives \(a^{p-1}\equiv1\pmod p\), and multiplying by \(a\) gives \(a^p\equiv a\pmod p\).
Useful when the base may depend on \(p\).