2

How to prove the following claim ?

$a$ is a fixed real number such that $n^a$ is an integer for all natural number $n$. Then , $a$ is an integer.

Please try this problem, I am having problems solving it.

Peter
  • 84,454
Arpan1729
  • 3,414
  • 2
    You just mentioned that $n$ is a natural number........ –  May 09 '17 at 15:47
  • 2
    may be we need to prove\disprove that $a$ is integer? – Mesmerized student May 09 '17 at 15:47
  • I'm voting to close this question because it's already been answered. – Parcly Taxel May 09 '17 at 15:49
  • 1
    Do you mean prove/disprove that $a$ is an integer? Other than that it would be helpful to provide some contex, where did you encounter this problem, your attempts on this problem etc. – kingW3 May 09 '17 at 15:49
  • @ParclyTaxel If it is a duplicater then close it as duplicate (and add the link to the answered question) – miracle173 May 09 '17 at 15:50
  • @miracle173 Metaphorically, I mean. In the comments. – Parcly Taxel May 09 '17 at 15:51
  • 2
    I haven't seen it in this site earlier, guys try it, its nice – Arpan1729 May 09 '17 at 15:53
  • 5
    Why all the close votes? Link to the duplicate if it exists please. And sure, the post is lacking own work, but the question itself is interesting. – TMM May 09 '17 at 16:08
  • @ArpanSadhukhan what a your question about this problem? From the title and a comment of yours I get the impression you might just be posing a question which you already know the answer to, and so you don't actually face a problem. This is not what our site is about (although it might be a fit for puzzling.se) – rschwieb May 09 '17 at 17:04
  • It is elementary that if $a$ is rational then $a$ is an integer. And the Gelfond theorem implies that $a$ cannot be an irrational algebraic number. But ....... – DanielWainfleet May 09 '17 at 18:12
  • My solution uses exponential sums. Anyone have a more elementary solution? – user397701 May 09 '17 at 21:20
  • @Kyle If memory serves, there is a classic elementary argument by Erdős, and that's why many others believe this to be a duplicate. I hope the dupe is found soon :). – Erick Wong May 10 '17 at 02:43
  • @Kyle My mistake, the Erdős result (at least the one I had in mind) was that if $f(n)$ is increasing and multiplicative then it must be $n^a$ for some $a>0$. – Erick Wong May 10 '17 at 02:56
  • 1
    Duplicate of https://math.stackexchange.com/questions/858893/if-any-integer-to-the-power-of-x-is-integer-must-x-be-integer – Erick Wong May 10 '17 at 03:07
  • a solution: https://mks.mff.cuni.cz/kalva/putnam/psoln/psol716.html – miracle173 May 10 '17 at 04:01

0 Answers0