NT-B1-M07-P021 Prime Divisors of \(2^p+1\)

For odd \(p\), use \(2^p\equiv2\pmod p\).

For \(p=2\), \(2^2+1=5\), not divisible by \(2\). Let \(p\) be an odd prime. By Fermat in the form \(a^p\equiv a\pmod p\), \(2^p\equiv2\pmod p\). Then \(2^p+1\equiv3\pmod p\). For \(p\mid2^p+1\), we need \(p\mid3\), so \(p=3\). Check: \(2^3+1=9\), divisible by \(3\). The answer is \(p=3\).

Olympiad move: the exponent itself is the prime modulus.