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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

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    $\begingroup$ Do you have examples, where you can show it works? $\endgroup$ – Milten Oct 31 at 11:54
  • $\begingroup$ sure, i'll get some up $\endgroup$ – Jakub Skop Oct 31 at 11:55
  • $\begingroup$ (3^3^3 +1)/(3^3+1) = 272342767321 $\endgroup$ – Jakub Skop Oct 31 at 11:56
  • $\begingroup$ (7^7+1)/(7+1) = 102943 $\endgroup$ – Jakub Skop Oct 31 at 11:57
  • $\begingroup$ (11^11 + 1)/(11 +1) = 23775972551 $\endgroup$ – Jakub Skop Oct 31 at 11:57
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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 $x$.

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  • $\begingroup$ Where did you get the factorisation from - I've never seen it before. $\endgroup$ – Jakub Skop Oct 31 at 12:24
  • $\begingroup$ Oops, I wrote it all wrong. I'll edit. $\endgroup$ – Milten Oct 31 at 12:48
  • $\begingroup$ Have you edited it; I can't tell $\endgroup$ – Jakub Skop Oct 31 at 12:54
  • $\begingroup$ Haha, yes I have. You can find it here under "Sum, odd exponent" with $E=x^m$, $F=1$. $\endgroup$ – Milten Oct 31 at 12:55
  • $\begingroup$ @JakubSkop I finally proved it :) $\endgroup$ – Milten Oct 31 at 13:15

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