Paper 4, Section I, H

Number Theory
Part II, 2015

Show that if 10n+110^{n}+1 is prime then nn must be a power of 2 . Now assuming nn is a power of 2 , show that if pp is a prime factor of 10n+110^{n}+1 then p1(mod2n)p \equiv 1(\bmod 2 n).

Explain the method of Fermat factorization, and use it to factor 104+110^{4}+1.