A repunit is a number consisting solely of ones (such as 11
or 11111).
Let us call p(n) a 10base integer represented by a
string of n ones, e.g. p(1)=1, p(5)=11111 etc.
Most of the repunit numbers are composite.
2, 19,23,317 are
the first four indices of prime repunits.
Prove: For a prime repunit p(n) to be prime, n has to
be prime.
As in an earlier comment, for n>1 the contrapositive "If n is composite, then p(n) is composite" is easier to show. (For n=1, note p(1) is not prime.)
Suppose n composite with a factorization xy, x>1, y>1. Then let q=p(y) with x1 zeroes placed between each pair of adjacent ones, and note p(n)=p(x)*q. x>1 implies p(x)>1, and y>1 implies q>p(y)>1. Thus p(n) is composite.
An example is n=6. If x=2, y=3 then p(x)=11, q=10101.

Posted by Gamer
on 20110110 18:45:40 