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I have tried to use reccurence demonstartion to show that :$(a+\frac{1}{n})^{n!}+(b+\frac{1}{n})^{n!}$ is integer number for $a, b$ are positive integers but i failed because i got a complicated formula .the last let me feel that is integer for some values of $n $, then My question here is :

How do I show that : $$(a+\frac{1}{n})^{n!}+(b+\frac{1}{n})^{n!}$$ is integer for finitely many positive integer $n$ for a postive integer $a$ and $b $ ?

Note: For some examples which i got in Wolfram alpha show that is very approximated to integers for $n=3$ as shown here with many values of $a$ and $b$ , and the motivation of this question is to know more about formula which produce integers and in the same time formula which produces primes !!!

Thank you for any help

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migrated from mathoverflow.net Nov 3 '16 at 0:15

This question came from our site for professional mathematicians.

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    $\begingroup$ Will this ever be an integer if $n>2$? Do you have any examples? $\endgroup$ – T. Amdeberhan Nov 2 '16 at 20:47
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    $\begingroup$ Again, do you any example of $a, b, n$ for which the given sum is an integer? $\endgroup$ – T. Amdeberhan Nov 2 '16 at 20:53
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    $\begingroup$ So this is what MO has evolved to ... $\endgroup$ – HeinrichD Nov 2 '16 at 21:16
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    $\begingroup$ @HeinrichD If you think this is a poor question, downvote it! Despite the grammar issues in the post there is a legitimately interesting question here - albeit one that might be worth migrating to math.SE rather than leaving here. $\endgroup$ – Steven Stadnicki Nov 2 '16 at 21:31
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    $\begingroup$ This can never be an integer for $n>2$ for elementary reasons. The question is indeed somehow legitimate but has nothing to do on MO, I think it should be moved to math.SE. (To prove it multiply by $n^{n!}$ and compute modulo $n$. If it were an integer it should give $0$, but the result is always $2$ modulo $n$, hence it cannot be an integer for $n>2$.) $\endgroup$ – Simon Henry Nov 2 '16 at 21:38

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