Wikipedia:Reference desk/Archives/Mathematics/2013 November 20

= November 20 =

Cyclic numbers
The article states that there are none in square bases. Why? Double sharp (talk) 05:37, 20 November 2013 (UTC)
 * If you look at how to form them then if the numerator $$b^{p-1}-1$$ is divisible by p, and b is the square of s then p will divide s so you don't get the longest length. Just factor the numerator and use Euler's theorem. Dmcq (talk) 10:22, 20 November 2013 (UTC)
 * Not quite... p is a prime that does not divide b, so p cannot divide s. However, if you expand the numerator: $$b^{p-1}-1 = s^{2(p-1)} -1 = (s^{p-1}-1)(s^{p-1}+1)$$, so any cyclic number of length p-1 in base b would have to be a multiple of a cyclic number of length p-1 in base s.  The multiplying factor, expressed in base p, is 1[0]1, where [0] is a number of 0s (actually ((p-1)/2)-1 0s), which means that the base b cyclic number is the base s cyclic number (expressed in base b) repeated twice. Since the number is cyclic in base s, all pemuations of that number in base s are multiples.  This will result in two sets of digits in base b (one for the even permutations, and one for the odd permuations), since two digits in s make up each digit in b. Note that the even and odd permuations are not necessarily the same as the even and odd multiples. MChesterMC (talk) 16:37, 20 November 2013 (UTC)
 * Sorry yes, I should have just given the theorem as a clue and left the rest as an exercise for the reader. Much less chance of me getting it wrong that way ;-) Dmcq (talk) 16:55, 20 November 2013 (UTC)