Can the 2008^th repunit be a perfect square in any integer base x ≥ 2?

Notice that x-1 times the repunit with n ones in base x equals x^(n+1)-1. So the repunit with n ones in base x equals x^(n+1)/(x-1)

Thus, we want to find x^2009-1/(x-1) a perfect square.

Posted by Gamer on 2007-10-07 00:17:55