Does the following statement hold? $$x\in \mathbb{R}^+ \text{and} \ 3^x, 5^x \in \mathbb{Z} \implies x \in \mathbb{Z}$$

In words:

If $x>0$ is a real number, and $3^x$ and $5^x$ are both integers, does that mean that $x$ is an integer?

This is a slightly modified form of another problem I was working on. A friend of mine claims this is a very hard problem. What do you think?

If one claims it is an open problem, can one show that this problem is equivalent to some other known open problem?

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    $\begingroup$ Related: mathoverflow.net/questions/17560/… $\endgroup$ – Ian Mateus Jan 21 '15 at 0:28
  • $\begingroup$ @IanMateus Thanks for the link, could you perhaps make a summary what it means for this question? (or put together a reasonable answer?) $\endgroup$ – VividD Jan 21 '15 at 0:54
  • $\begingroup$ @VividD it means this is an open question in the field of mathematics. $\endgroup$ – Angad Jan 21 '15 at 1:08
  • $\begingroup$ @Angad, this is not so obvious. $\endgroup$ – VividD Jan 21 '15 at 1:17
  • $\begingroup$ The case is actually not listed. Only that the case with more information ($2^x$) needs bad-a** algebra, and that a similiar case ($2^x$ and $3^x$) is open. $\endgroup$ – mvw Jan 21 '15 at 1:18

This is probably an open question, as the related problem with $2^x$ and $3^x$ is open. Today, it is known that if $2^x$, $3^x$ and $5^x$ are integers, then $x$ is integer as well--it follows from the six exponentials theorem in transcendental number theory.

I cannot confirm whether the $3^x$, $5^x$ case follows from the four exponentials conjecture, as I do not know the field; so I would be glad if someone could.

  • $\begingroup$ Fair enough, thanks! I believe so too, now that I read more in the linked texts. $\endgroup$ – VividD Jan 21 '15 at 1:30
  • $\begingroup$ Hope somebody with knowledge in that area will kick in. $\endgroup$ – VividD Jan 21 '15 at 1:32
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    $\begingroup$ source material on six exponentials theorem math.stackexchange.com/questions/1087841/… from Lang's book $\endgroup$ – Will Jagy Jan 21 '15 at 1:46

I would say yes. If we assume that $x\not\in\mathbb{Z}$ we can write it as $n+\alpha$ where $n\in\mathbb{Z}$ and $\alpha\in (0,1)$ then $$3^x=3^{n+\alpha}=3^n\cdot 3^{\alpha}$$

We know that $3^n$ is an integer and if $3^x$ is an integer too then $3^\alpha$ must be one too.

Now, as $\alpha\in(0,1)$ we can write it as $\frac{1}{\beta}$ where $\beta>0$ so we get that


which is definitely not an integer which is a contradiction so $x$ must be an integer.

Maybe I'm missing something in the original question but I don't see how the $5^x$ changes anything.

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    $\begingroup$ Well, $\log_3{2}$ is a real number between $0$ and $1$, but $3^{\log_3{2}}=2\in\mathbb{Z}$. $\endgroup$ – ryagami Jan 21 '15 at 0:28
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    $\begingroup$ Have a look at a plot of $\sqrt[\beta]{3}$ $\endgroup$ – mvw Jan 21 '15 at 0:31
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    $\begingroup$ Also, there's no reason $3^{\alpha}$ must be an integer (as you say in the second line). For example, if $3^{\alpha} = \frac{2}{3}$ and $n=2$, then $3^n\cdot 3^\alpha = 9\cdot\frac{2}{3} = 6 = 3^x$ is an integer. $\endgroup$ – user88319 Jan 21 '15 at 0:39
  • $\begingroup$ How can I be so wrong :( $\endgroup$ – fcortes Jan 21 '15 at 21:41

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