Can the 2008
th repunit be a perfect square in any integer base x ≥ 2?
(In reply to
Repunits by Gamer)
"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)"
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).
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Posted by Charlie
on 2007-10-07 02:35:47 |