A repunit is a number consisting solely of ones (such as 11 or 11111).
Let us call p(n) a 10-base 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.

(In reply to re: Another proof - not a proof by Ady TZIDON)

I don't understand the problem. In this case, "q=p(y) with x-1 zeroes placed between each pair of adjacent ones" results in q depending on x and y.

If you set x=3, y=2, then p(x)=111 as you mention. However q=1001 in this case, as p(y)=11 and placing x-1=2 zeroes between each pair of ones results in 1001.

Posted by Gamer on 2011-01-11 14:16:32