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


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

This question came from our site for professional mathematicians.

  • 1
    $\begingroup$ Will this ever be an integer if $n>2$? Do you have any examples? $\endgroup$ – T. Amdeberhan Nov 2 '16 at 20:47
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ So this is what MO has evolved to ... $\endgroup$ – HeinrichD Nov 2 '16 at 21:16
  • 1
    $\begingroup$ @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
  • 3
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.