Problem
NT-B1-M12-P022 Set 5. A Prime Divisor Condition
#22
★★★★☆ Level 4 of 5
Find all primes \(p\) such that \(p\mid2^p+1\).
For an odd prime, use Fermat's little theorem.
For \(p=2\), \(2^2+1=5\), not divisible by \(2\). Let \(p\) be odd. By Fermat's little theorem, \(2^p\equiv2\pmod p\). Then \(2^p+1\equiv3\pmod p\). Thus \(p\mid3\), so \(p=3\). Check: \(3\mid9\).
Short problem with a strong method choice.