More interesting than I at first thought.
A simple approach is to consider divisibility by 2. Clearly if n^5+5 is even, then (n+1)^5+5 is odd and vice versa. And the same with 3,5,7, and many other possible prime factors. So (one might wrongly think) n^5+5 is periodic with length n, problem solved.
But divisibility by 11 gives no integer solutions. The same is true for {11, 41, 61, 71, 101,...}. Easy, (one might again wrongly think) just add a proviso that either n^5+5 is periodic with length n, or there is no period. But this isn't true either; n = {31,191, 251, 271, 601, 641, 761...} produce multiple periods (e.g 31 has 31k+3, 31k+6,31k+12, 31k+17, and 31k+24. And compound multiples of these numbers work too (31*2, 31*3,31*7 etc also have multiple periods).
I don't think it's in the least likely that one of these numbers, or its compounds, is going to have 2 successive periods of the form (ak+x),(ak+x+1) but I don't see an obvious way to show it.

Posted by broll
on 20120822 12:41:47 