Is 10101...01 composite in every base?

Question

Given all numbers in base b, where b is an integer ≥ 2, that consist of alternating 1s and 0s. So, our numbers include 101, 10101, 1010101, etc.

Which of these numbers can be prime?

Hmm... there are 32 people in my "My Fans" section, some of whom I've seen answer questions way harder than this. Surely someone wants Best Answer for this. :)

If the "any base" requirement is confusing the issue, I'll accept an answer for base 10 numbers only. Actually, if you solve it for base 10, the way to generalize may become apparent.

Answer 1

Let x = base. Then we're talking about numbers in the form

1 + x^2 + x^4 + x^6 + ... etc.

Except for 1 + x^2, these polynomials can always be factorized, having at least for one factor either 1 + x^2 or 1 - x + x^2, the first in the case of an even number of 1s, and the second for odd number of 1s. So, this question reduces to whether or not the number 101 is composite. We know it's prime in base 10, but not for odd bases, since they will always generate an even number (just expand (2n+1)^2 + 1). So, finally, are there any even bases for which it's composite? If not, what's the rule? This is the same as asking for which value of n is the number 4n^2 + 1 a prime, and this is a far harder problem than the original question. I'll get back to you on this one when I'm ready for the Fields Prize.

Answer 2

Your numbers are all of the form Sum_{n=0}^{k} a^{2n} = Sum_{n=0}^{k} (a^2)^n which is a geometric sum that simplifies to (1 - a^{2(k+1)})/(1-a^2). I'm not sure if this helps or not, but it's a start, as this is a compact symbolic representation of your question.