There I was, just messing around with tetration, when I stumbled across this -
$(x^x +1)/(x+1)$ = integer (for odd integer values of x)
Playing some more with this it seems (not entirely sure as tetration quickly becomes to hard to compute) -
$({{{^n}^+}^1}x + 1)$/$(^nx + 1)$ --> Integer
(also for odd integer values of x)
I am aware of that this is to do with the factors of $x^x +1$ ,but can anyone give a full deeper visceral explanation to this? (If it is true - if not then tell me)
Thanks
We have the factorisation: $$ x^{k}+1 = (x+1)(1-x+x^{2}-\cdots +x^{(k-1)}) $$ for any odd natural number $k$.
$^n x$ divides $^{n+1}x$, since they are both just a bunch of $x$-factors. So we can write $^{n+1}x = k\cdot{}^n x$ for some odd $k$ (since $^{n+1}x$ is odd, $k$ has to be odd). Then we can factorise: $$ {}^{n+1}x + 1 = x^{{}^nx} + 1 = x^{k\cdot ({}^{n-1}x)} + 1 \\ = ({}^n x)^k+1 = ({}^n x+1)\left(1-({}^n x)^{({}^n x)}+({}^n x)^{2({}^n x)}-\cdots +({}^n x)^{(k-1)({}^n x)}\right) $$ which proves the result.
BONUS:
The factorisation we used is a variation of the following, which works for any $k$: $$ x^{k}-1 = (x-1)(1+x+x^{2}+\cdots +x^{(k-1)}) $$ Therefore we can conclude by the same method as above that $({}^n x-1)$ divides $({}^{n+1}x - 1)$ for any $x$.