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I am a high school student and came across this problem on Number Theory.

(1) Prove that for any given positive integer $n$, there exists a set $S \subseteq \mathbb{N^{*}}$, such that for all $1\le k \le n$, any subset $T$ of $S$ of size $k$, satisfies $\displaystyle \prod_{x\in T}x=t_x^k,t_x \in \mathbb{N}$.

(2) Prove that it is impossible to find an infitite set $S \subseteq \mathbb{N^{*}}$, such that for all positive integer $k$, any subset $T$ of $S$ of size $k$ satisfies $\displaystyle \prod_{x\in T}x=t_x^k,t_x \in \mathbb{N}$.

Note that $t_x$ can be different for diffrent $x$. Or as in Chinese, 任意 $k$ 元子集之积都是 $k$ 次方幂。

I can easily give a construction of the first question, which is to arbitrarily choose $n$ integers $a_1, a_2, \ldots, a_n$, and let $S=\{a_1^{n!}, a_2^{n!}, \ldots, a_n^{n!}\}$. However, I just cannot understand why I can find such a set of arbitrary size but cannot find such a set of infinite size.

Can anyone explain why to me in a high-schooler-understandable way?

update1: Chinese statement.

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  • $\begingroup$ I don't think it's the size of $S$ that makes the big difference (you could easily modify your solutino to (1) to let $S$ be infinite), but rather than in (2) one needs to handle infinitely many exponents in $t^k$. (The assumption that $S$ be infinite in (2) isn't necessary to state because it's obvious that a finite $S$ doesn't contain subsets of size $k$ for all $k\in \Bbb N$.) $\endgroup$ Aug 28, 2021 at 3:39
  • $\begingroup$ That being said, I'm not sure (2) is accurate: it seems that the example $S=\{1^1, \; 2^2, 3^2,\; 4^3, 5^3, 6^3, \; 7^4, 8^4, 9^4, 10^4, \; \dots \}$ works. $\endgroup$ Aug 28, 2021 at 3:41
  • $\begingroup$ Sorry, I mis-translated! It should be any subset of size $k$ $\endgroup$
    – Lily White
    Aug 28, 2021 at 3:43
  • $\begingroup$ @GregMartin I have edited the question a bit and clarified something. It's hard to write mathematical-English really. $\endgroup$
    – Lily White
    Aug 28, 2021 at 3:49
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    $\begingroup$ Technically, the property is true for an infinite set, if you still only have the property for $1\leq k\leq n.$ It’s only when you cosider the size of $S$ and the maximum for $k$ as the same that it fails. $\endgroup$ Aug 28, 2021 at 3:49

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