# If all the numbers $(1^\alpha,\,2^\alpha,\,3^\alpha,\,\dotsc)$ are integer, then $\alpha$ is an integer.

A theorem of Siegel asserts that

If $\beta>0$ and $2^\beta,\,3^\beta,\,5^\beta$ are integers, then $\beta$ is an integer.

The following result is a beautiful consequence of this theorem

If $\beta$ is a positive number such that $1^\beta,\,2^\beta,\,3^\beta,\,\dotsc$ are integers, then $\beta$ is itself is an integer.

I'm looking for a proof of this result.

Note. This result appeared as a problem in the 1972 Putnam Prize competition, and not one of more than 2000 university student competitors gave a solution; the solution, though not hard, could well elude even a professional mathematician for several hours (or days).

• I fixed typos in the title. Is it OK ? Apr 26 '15 at 14:40
• This is a restatement of Problem A6 from the 1971 Putnam; a solution appears here. Apr 26 '15 at 14:43
• We may notice that it is a pretty straightforward consequence of Baker's theorem (en.wikipedia.org/wiki/Baker%27s_theorem). Apr 26 '15 at 14:44
• @vadim123: oh, now I see. The solution through finite differences is very nice and tricky! Apr 26 '15 at 14:47
• There is a solution for the problem with using finite differences, however the generalisation of the Siegel theorem for two bases is open.
– k1.M
Apr 27 '15 at 8:17

Well, k1.M, here an extension of important and beautiful Siegel Theorem (I did not know). It is clear $\beta$ must be integer or irrational and by another theorem of Siegel $\beta$ must be trascendental. I guess the statement goes for any three coprime integer x, y, z the triple 2, 3, 5 having been chosen for smallest.Thus the true spirit of the statement of Siegel would be "all trascendental as exponent of an integer can gives an integer at most twice".