Problem
NT-B1-M01-P023 Greatest Divisor of \(n^5-n\)
#23
★★★★☆ Level 4 of 5
Find the greatest positive \(m\) such that \(m\mid n^5-n\) for every integer \(n\).
Prove divisibility by \(2\), \(3\), and \(5\), then substitute \(n=2\).
Modulo \(2\), always \(n^5\equiv n\), so \(2\mid n^5-n\). Modulo \(3\), checking residues \(0,1,2\) gives \(n^5\equiv n\). Modulo \(5\), for residues \(0,\pm1,\pm2\), we also get \(n^5\equiv n\). Hence \(30=2\cdot3\cdot5\mid n^5-n\) for all \(n\). On the other hand, at \(n=2\), \(2^5-2=30\), so every universal divisor \(m\) must divide \(30\). Therefore the greatest \(m\) is \(30\).
Mixes divisibility, residues, and the maximality idea.