For a prime p with p-1 a power of two, a zero geometric sum of odd length forces p to divide the length

Let p be a prime whose predecessor p-1 is a power of two, and let t be a natural number with p not dividing t. If p divides the geometric sum 1 + t + t^2 + ... + t^(2e) of odd length 2e+1, then p divides 2e+1. If t is congruent to 1 modulo p the sum is congruent to 2e+1, so p divides the length. Otherwise the multiplicative order of t modulo p divides the odd number 2e+1 and is therefore odd, but a cyclic group of order a power of two has no nontrivial element of odd order, a contradiction. This is the order-theoretic obstruction behind the k=5 second-Dris-equation residual: the demand that the p-adic valuation of sigma(m^2) be at least 5 cannot be met unless some prime factor of m occurs to an exponent e with p dividing 2e+1.