Problem
NT-B2-M09-P013 Composite Triples
#13
★★★★☆ Level 4 of 5
Prove that there are infinitely many \(n\) such that \(n\), \(n+2\), and \(n+6\) are all composite.
Assign distinct prime divisors and then take large solutions.
Require \(n\equiv 0\pmod 5\), \(n+2\equiv 0\pmod 7\), \(n+6\equiv 0\pmod {11}\). This is \(n\equiv 0\pmod 5\), \(n\equiv 5\pmod 7\), \(n\equiv 5\pmod {11}\). The moduli are coprime, so there is a solution \(n_0\) modulo \(385\). All \(n=n_0+385t\) have the same divisibility properties. For sufficiently large \(t\), the numbers \(n,n+2,n+6\) are larger than their assigned divisors, hence composite. There are infinitely many.
An olympiad-style construction resembling a way to destroy prime triples.