Problem
NT-B2-M04-P013 All Bases
#13
★★★★☆ Level 4 of 5
Let \(p\) be prime. Prove that \(p\mid a^{p+1}-a^2\) for every integer \(a\).
Factor \(a\), or use \(a^p\equiv a\).
By Fermat's form, \(a^p\equiv a\pmod p\). Multiplying by \(a\), we get \(a^{p+1}\equiv a^2\pmod p\). Hence \(p\mid a^{p+1}-a^2\).
Trains flexible use of \(a^p\equiv a\).