Problem
NT-B2-M12-P015 Composite Terms \(2^{2^n}+3\)
#15
★★★★☆ Level 4 of 5
Prove that the sequence \(2^{2^n}+3\), \(n=1,2,\ldots\), contains infinitely many composite numbers.
1001 Problems in Classical Number Theory (method inspiration) · Problem 355
Try to find a prime modulus \(p\) for which \(-3\) is a power of \(2\).
Modulo \(19\), \(2^4=16\equiv-3\). Thus it is enough to have \(2^n\equiv4\pmod{18}\), because the order of \(2\) modulo \(19\) divides \(18\). The residues \(2^n\pmod{18}\) are periodic, and \(2^n\equiv4\pmod{18}\) when \(n\equiv2\pmod6\). Hence for all \(n\equiv2\pmod6\), the number \(2^{2^n}+3\) is divisible by \(19\). For \(n>2\), it is greater than \(19\), hence composite. There are infinitely many such \(n\).
This problem nicely shows periodicity of an exponent inside an exponent.