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

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.

• $m$ is the number of elements in the subset of $S_n$. Oct 18, 2013 at 8:42
• $n$ is just the number of elements in $S_n$. Your comment doesn't make any sense. If there us a simple proof, I encourage you to post it. The simpler the better. Saying that something is 'obviously false', and claiming that the proof is 'simple' is a bit too reminiscent of Fermat, I think. Oct 20, 2013 at 15:32
• I just think you could have worded the question better: just say, "for any finite subset of $S$, the geometric mean of all its elements is an integer". No need for confusing $m$'s or $n$'s. It is intuitively compelling that such an $S$ can't exist, but I admit I haven't proven it. I probably shouldn't have written "obvious". Someone gave a four-sentence proof that no one has objected to, so it is probably correct. If you disagree with that answer or you don't follow it, I suggest you leave the answerer a comment. Oct 21, 2013 at 16:27
• Please see the expanded answer, which I found easier to read. This is a nice problem. My initial reaction was that it should be easy, but it's harder to prove than I expected. Oct 21, 2013 at 18:57

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.

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 $3$.

• What's "ord"? I am not a number theory expert, and I'd like to understand your answer. Also, $S$ is the entire, infinite set, so using "$S$" to refer to an $m+1$-element sequence may be a bad idea. Oct 21, 2013 at 16:31
• @StefanSmith I changed $S$ to $T$. $\operatorname{ord}_p a$ is the largest number of powers of $p$ that divide $a$. This is often called the "order of $a$ at $p$", for the same reason that counting the number of factors of $x$ in a polynomial $P(x)$ tells you the order of vanishing at $x=0$. Oct 21, 2013 at 18:46
• @StefanSmith Oh, I also clarified the proof substantially. Oct 21, 2013 at 18:46
• Thanks. It might be a good idea to define "ord" within your answer as well as in the comment (it might increase the probability of the OP accepting your answer). Nice proof. Oct 21, 2013 at 18:55