Let $x,y>1$ be coprime integers and $g>0$ a real number such that $g^x,g^y$ are both integers. Is it true that $g\in\mathbb N$

elementary-number-theorygcd-and-lcminteger-latticesprime factorizationproof-writing

Let:

  • $x, y\ $ be coprime integers greater than $1$
  • $g \in \mathbb{R}^+$
  • $g_,^x \ g^y \in \mathbb{N}$

Proposition: $g \in \mathbb{N}$

I have not managed to prove it. Via the fundamental theorem of arithmetic, I have proved (at least I believe so) that if $g$ is rational, it is an integer. However, I am not sure about the case of irrational $g$; I believe such irrational $g$ can exist. Can anyone (dis-)prove this?

My intuition behind it is this (I'm only using $x$ but the same applies to $y$): for $g^x$ to be integer, $g$ must be the $x^\text{th}$ root of $r_x = g^x$, i.e. $g = \sqrt[x]{r_x}$. We can construct the divisor lattice of $r_x$ and its "divisor cuboid": a generalized Hasse diagram of divisors of $r_x$, in wich the divisors are equally spaced, and which is continuous, viz. there are real numbers between the integer divisors. Now take the vector spanning the distance from the origin (the number $1$) to $r_x$ in the divisor cuboid, and scale down this vector by factor $x$ to obtain $\sqrt[x]{r_x} = g$. To allow for a $y$ and $r_y$ in all possible cases of $r_x$, we must extend the divisor cuboid to all the primes. Then there are unconutably many points in the divisor cuboid in which the value equals $g$, since we can express $g$ as a real power of any single prime (or a product thereof). So it may be possible to find several such $n \in \mathbb{N}\:$, which are mutually prime and for which $g^n \in \mathbb{N}$. But I am not even fully convinced of this myself; let alone consider this a proof that could convince someone else.

Best Answer

If $x$ and $y$ are coprime then there exist integers $a$ and $b$ such that $ax+by=1$. So if $g^x$ and $g^y$ are both natural numbers, we see that $g^{ax+by} = g$ is a rational number. So $g$ is rational. Say $g=p/q$ where $p$ and $q$ are relatively prime.

Now we ahve $g^x = p^x/q^x = n$ where $n$ is a natural number. Therefore $q$ divides $p^x$. But since $p$ and $q$ are relatively prime, we must have that $q$ divides $p$, meaning that the only possibility is that $q=1$. This means that $g$ is a natural number.

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