A number like 11, 111, or 1111 i.e. a number containing only the digit 1
is called
repunit.
List all base-8 prime repunits.
The base is 8, which makes me think the proof will be based off of a polynomial factorization.
Create a set of polynomials P_n(x) = x^(3n) + x^(3n-3) + ... + x^3 + 1. Then each base 8 repunit is P_n(2) for a repunit of n+1 digits.
Playing with set of polynomials, it seems that every P_n(x) is factorable with a factor of the form x^n + x^(n-1) + ... + x + 1.
If I assume this is true then the proof falls into place by simply factoring each polynomial P_n(x) and picking up the oddball cases where one of the factors turns out to evaluate to 1.
Almost a proof
