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suppose that $a$ is a postive real number such that all numbers $$1^a,2^a,3^a,\cdots$$ are integers,Prove that $a$ is also integer

This problem is 1971 Putnam competition,and the Official answer is give Using Lagrangian mean value theorem in higher mathematics, see answer,So there is no elementary practice, middle school students can do?

math110
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    Did you get some progress? For example, is it clear that $a$ cannot be irrational using "middle-school" methods, for example? Unfortunately, I am not sure that you can do this in general, because how will you handle transcendental $a$ like logs and so on, which don't even fall in the middle-school domain? For example, $2^{\log_2 3} = 3$ is an integer, but the exponent is not even definable, or cannot be worked with using middle-school methods. – Sarvesh Ravichandran Iyer Mar 28 '19 at 05:07
  • Also, please copy the official solution itself onto your question for completeness. (Follow on : thank you for doing that! As I mentioned earlier, it is difficult to handle log-type exponents without appealing to calculus. For one,we cannot even work with $a^b$ without using calculus). – Sarvesh Ravichandran Iyer Mar 28 '19 at 05:12
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    The Putnam is, and has always been, a contest for university students. Calculus knowledge is expected for all test takers, and is fair game for even the easiest Putnam problems. – jmerry Mar 28 '19 at 05:26
  • But it is said can use middle students methods to solve it,so I ask it – math110 Mar 28 '19 at 05:28
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    @inequality Who stated "it is said" that middle students are able to solve it? – John Omielan Mar 28 '19 at 05:52
  • FYI, I think "If $2^{a}, 3^{a}$ are integers, then $a$ is also an integer" is an open problem. – Seewoo Lee Apr 01 '19 at 01:21
  • @SeewooLee Note the OP provided a link to a solution at Putnam 1971/A6 solution. – John Omielan Apr 01 '19 at 01:41
  • @JohnOmielan I know, I just want to say that the stronger problem is an open problem. – Seewoo Lee Apr 01 '19 at 01:46
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    @inequality Please refer me to whoever called this a middle school problem! It is astounding that people should even expect to know calculus at the middle school level. – Sarvesh Ravichandran Iyer Apr 01 '19 at 07:59
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    Linked. – Alex Ravsky Apr 03 '19 at 05:03

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This is a partial answer: The problem can be reformulated as follow, let $\epsilon=\{a\}=a-\lfloor a \rfloor$ be the fractional part of the real number $a$: $$\mbox{Given} \quad 0\leq \epsilon < 1, \quad n^\epsilon \in \mathbb{Q} ,\quad \forall n \in \mathbb{N} \implies \epsilon =0 $$ One can show that $\epsilon$ can not be rational since for any prime number $p$ and positive integers $k < m$ such that $\gcd(k,m)=1$ we can easily show that $\sqrt[m]{p^k}$ is irrational using Fundamental Theorem of Arithmetic and same argument as in proving $\sqrt{2}$ is irrational. The remaining cases are the most difficult to get "elementary proof" and those cases are when $\epsilon$ is irrational.

HassanB
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