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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.

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  • $\begingroup$ $m$ is the number of elements in the subset of $S_n$. $\endgroup$
    – Gerard
    Oct 18, 2013 at 8:42
  • $\begingroup$ $n$ 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. $\endgroup$
    – Gerard
    Oct 20, 2013 at 15:32
  • $\begingroup$ 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. $\endgroup$ Oct 21, 2013 at 16:27
  • $\begingroup$ 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. $\endgroup$ Oct 21, 2013 at 18:57

1 Answer 1

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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 $3$.

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  • $\begingroup$ 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. $\endgroup$ Oct 21, 2013 at 16:31
  • $\begingroup$ @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 $x=0$. $\endgroup$
    – Slade
    Oct 21, 2013 at 18:46
  • $\begingroup$ @StefanSmith Oh, I also clarified the proof substantially. $\endgroup$
    – Slade
    Oct 21, 2013 at 18:46
  • $\begingroup$ 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. $\endgroup$ Oct 21, 2013 at 18:55

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