Problem
NT-B2-M11-P020 Prime Divisors of a Prime-Length Repunit
#20
★★★★★ Level 5 of 5
Let \(q\) be an odd prime and \(R_q=\frac{10^q-1}{9}\). Prove that every prime divisor \(p\ne3\) of \(R_q\) satisfies \(p\equiv1\pmod q\).
Consider the order of \(10\) modulo \(p\). It divides \(q\), but it is not \(1\).
From \(p\mid R_q\), we get \(10^q\equiv1\pmod p\), because \(9R_q=10^q-1\). Since \(p\ne3\), there is no exceptional division by \(9\); also \(p\ne2,5\). Let \(d\) be the order of \(10\) modulo \(p\). Then \(d\mid q\). Since \(q\) is prime, \(d=1\) or \(d=q\).
If \(d=1\), then \(10\equiv1\pmod p\), so \(p\mid9\), hence \(p=3\), which is excluded. Therefore \(d=q\). By the order theorem, \(d\mid p-1\), so \(q\mid p-1\), i.e. \(p\equiv1\pmod q\).
A strong closing problem: it combines repunits, order, and prime divisors.