Problem
NT-B2-M03-P016 A Prime Divisor of \(3^4+1\)
#16
★★★★☆ Level 4 of 5
Find all primes \(p\) such that \(p\mid3^4+1\).
First compute the number, then explain the restriction by order.
\(3^4+1=82=2\cdot41\). Hence \(p=2\) or \(p=41\). For the method: if \(p\ne2\) divides \(3^4+1\), then \(3^4\equiv-1\), so the order of \(3\) modulo \(p\) is \(8\), and \(8\mid p-1\). Among the divisors of \(82\), this holds only for \(41\).
The problem shows how order explains the factorisation.