11
$\begingroup$

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?

$\endgroup$
7
  • 6
    $\begingroup$ 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
4
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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
  • 1
    $\begingroup$ source material on six exponentials theorem math.stackexchange.com/questions/1087841/… from Lang's book $\endgroup$ – Will Jagy Jan 21 '15 at 1:46
0
$\begingroup$

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

$$3^{\alpha}=\sqrt[\beta]{3}$$

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.

$\endgroup$
4
  • 9
    $\begingroup$ 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
  • 4
    $\begingroup$ Have a look at a plot of $\sqrt[\beta]{3}$ $\endgroup$ – mvw Jan 21 '15 at 0:31
  • 3
    $\begingroup$ 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

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.