The easiest way to approach this is to look for n as a factor in the (n-1)th or earlier terms of x since the RHS is multiplied by x(n-1) and then divided by n, to produce xn. Then we see immediately that each successive 2(2n-1) incorporates a 'bow-wave' of numbers whose factors are potential divisors of the subsequent values of x, in ample time for them to be 'knocked out' again by being divided out by some n, so that there are always sufficient matching factors available for the division. For example, the factor 7 appears when n=4, well in advance of n = 7; and again at n = 11 (and every subsequent n = 7k-3) in good time to divide out the n's having 7 as a factor.
Accordingly, xn is always an integer.
Edited on October 4, 2010, 1:44 pm
Posted by broll
on 2010-10-04 13:35:44