Eötvös 1940/2 solution

44th Eötvös 1940

Problem 2

m and n are different positive integers. Show that 2^{2^m} + 1 and 2^2ⁿ + 1 are coprime.

Solution

wlog m < n. Suppose p divides both 2^{2^m} + 1 and 2^2ⁿ + 1. Then since they are both odd, p must be odd. It must also divide their difference 2^{2^m}(2^{2^n-m} - 1) and hence 2^{2^n-m} - 1. But (2^{2^n-m} - 1)(2^{2^n-m} + 1) = (2^{2^n-m+1} - 1), so p also divides (2^{2^n-m+1} - 1) and so by a trivial induction (2^2ⁿ - 1). But it divides (2^2ⁿ + 1), so it must divide their difference 2. But p is odd, so they have no common factors.

44th Eötvös 1940