Can the 2008
^{th} repunit be a perfect square in any integer base x ≥ 2?
(In reply to
Repunits by Gamer)
"Notice that x1 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)/(x1)"
But then, in base x = 10, with one 1, this gives 1 = (10^2)/9. With two 1's, 11 = (10^3) / 9.
Actually, in base x, a repunit of n 1's = (x^n  1) / (x  1).

Posted by Charlie
on 20071007 02:35:47 