Does there exist an infinite set of integers such that the geometric mean of any of its subsets is an integer?

Question

Is there an infinite set $S$ such that for any subset with $m$ integers the geometric mean is also an integer?

We can always find a set, $S_n$ with $n$ elements which satisfies the given requirement:

$$S_n = \left\{a^{n!}_1, a^{n!}_2, a^{n!}_3, \dots, a^{n!}_n\right\}$$

for distinct integers $a_1, a_2, \dots, a_n \neq 1$.

Consider a subset of $S_n$, of $m$ elements denoted by $S'_m$ where $m \leq n$.

$$S'_m = \left\{a'^{n!}_1, a'^{n!}_2, \dots, a'^{n!}_m\right\}$$

The GM of this equals:

$$\left(a'^{n!}_1a'^{n!}_2\dots a'^{n!}_m\right)^{\frac1m}$$

$$ = \left(a'^{\frac{n!}{m}}_1 a'^{\frac{n!}{m}}_2 a'^{\frac{n!}{m}}_3 \dots a'^{\frac{n!}{m}}_m\right)$$

Since $m|n!$, all powers are integers. Therefore, the GM is an integer.

Now, consider the case as $n \rightarrow \infty$:

This implies, $n!$ approaches $\infty$. Each of the elements of the set approaches $\infty$. Can there be an infinite number of integers equal to infinity in a set? I'm confused.

Answer 1

No, it's impossible.

Definition: $\operatorname{ord}_p x$ is the number of powers of $p$ that divide $x$.

Consider any prime $p$. If $T$ is an $m+1$-element sequence such that the geometric mean of any $m$ elements is an integer, then, for any $x,y\in T$, we have $\operatorname{ord}_p x \equiv \operatorname{ord}_p y\ (\operatorname{mod} m)$.

It follows that, for any $x,y\in S$, $\operatorname{ord}_p x - \operatorname{ord}_p y $ is divisible by any positive integer $m$, so we must have $\operatorname{ord}_p x = \operatorname{ord}_p y$. Since this holds for all primes $p$, we must have $x=\pm y$.

But we can't have $x=-y$ (unless they're both zero), because then their geometric mean would not be defined. So $x=y$, and we conclude that our sequence is constant.

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Here is an example of the above proof in action:

Let $T=\{a,b,c,d\}$, and $p=5$. Then, write: $a=a'\cdot 5^u$, $b=b'\cdot 5^v$, $c=c'\cdot 5^w$, and $d=d'\cdot 5^l$, where none of $a',b',c',d'$ are divisible by $5$.

Then the geometric mean of $a$, $b$, and $c$ is $\sqrt[3]{a'b'c'}\cdot 5^{\frac{u+v+w}{3}}$, so $u+v+w$ is divisible by $3$, since we have assumed that the geometric mean is an integer.

Similarly, $u+v+l$ is divisible by $3$, so $w-l = (u+v+w) - (u+v+l)$ is divisible by $3$. In other words, $\operatorname{ord}_5 c - \operatorname{ord}_5 d $ is divisible by $3$.