Problem
NT-B2-M03-P020 Hidden Order
#20
★★★★★ Level 5 of 5
Let \(p\) be an odd prime, \(p\nmid a\), and \(p\mid a^6-1\), but \(p\nmid a^3-1\) and \(p\nmid a^2-1\). Prove that \(p\equiv1\pmod6\).
Determine the order of \(a\) modulo \(p\).
Let \(d=\operatorname{ord}_p(a)\). From \(a^6\equiv1\pmod p\), we get \(d\mid6\). The conditions \(p\nmid a^3-1\) and \(p\nmid a^2-1\) mean \(d\nmid3\) and \(d\nmid2\). Among the divisors of \(6\), only \(6\) divides neither \(2\) nor \(3\). Hence \(d=6\). The order modulo a prime divides \(p-1\), so \(6\mid p-1\), meaning \(p\equiv1\pmod6\).
The final problem checks whether the student can reconstruct the order from negative conditions.