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How to show that these sets are nonempty (here $\mid $ means "divides")?

Here N is an arbitrary large integer and q is some fixed integer.

$R = \lbrace k \in {\mathbb N}:(kN\mid k!) \wedge ((k - 1)N\mid k!) \wedge \cdots \wedge (N\mid k!) \wedge (k > Nq)\rbrace$

$S = \lbrace k \in {\mathbb N}:({(2k - 1)^2}N\mid k!) \wedge ({(2k - 3)^2}N\mid k!) \wedge \ldots \wedge (N\mid k!) \wedge (k > Nq)\rbrace$

$T = \lbrace k \in {\mathbb N}:({k^5}N\mid k!) \wedge ({(k - 1)^5}N\mid k!) \wedge \ldots \wedge (N\mid k!) \wedge (k > Nq)\rbrace$

They exist by the axiom schema of separation, but how do I determine which $k$ to choose so that it satisfies all the properties? Is there a general approach?

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    I don't think it is true for $S$. Let $p$ be a prime between $k$ and $2k$. (Such a $p$ exists by Betrand's Postulate.) Then $p\not\mid k!$, so definitely $p^2N\not\mid k!$2012-07-23
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    Similarly, I don't think it is true that $T$ is non-empty, since if $p$ is a prime between $k/2$ and $k$ then $p^2\not\mid k!$ so definitely $p^5N\not\mid k!$2012-07-23
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    That does not mean ${p^2}\nmid k!$, does it?2012-07-23
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    What does not mean it? If $p\not\mid k!$ then $p^2\not\mid k!$. @glebovg2012-07-23
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    Choose $k > {p^2}$, ${p^2}$ is an integer so ${p^2}\mid k!$.2012-07-23
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    But I'm pointing out that once you've asserted some $k$, I can find a $p$ which proves that $k$ is not in the set. You can't then change $k$.2012-07-23
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    For example, I can show to you that $k=500$ is not in the $S$, because $503\not\mid 500!$, and hence $503^2N\not\mid 500!$. And, given any $k$ you choose, I can likewise find an odd prime $k.2012-07-23
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    Essentially, I want to find a specific k, which would satisfy all the properties. Perhaps, there exists a unique k.2012-07-23
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    Yes, some sets are involved, and the axiom schema of separation was mentioned. That does not make a set theory question, though.2012-07-23
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    If no such k exists, there must be a way to prove it.2012-07-23
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    @glebovg I did prove it. I proved for each $k$, $k\not\in S$2012-07-23
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    I do not understand. You only used some examples. For k large, k! is much larger than ${(2k - 1)^2}N$, ${(2k - 3)^2}N$, etc. and they are all integers, so perhaps there is some k for which all of the above divide k!.2012-07-23
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    @Thomas Andrews If, indeed, you did prove it, could you provide a formal proof?2012-07-23
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    Let $k$ be a natural number. If $k=1$ then it is not true that $k>Nq$, so $k\not\in S$. If $k>1$, then there is a prime $p$ such that $k. Since $p\not\mid k!$, $p^2N\not\mid k!$, so $k\not\in S$. Hence, $\forall k\in \mathbb N: k\not\in S$2012-07-23
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    Why does p not divide k!? ${(2k - 1)^2}$, ${(2k - 3)^2}$, etc. may all not be prime.2012-07-23
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    @Thomas Andrews I am sorry, I do not see how your argument about primes is relevant. Can you elaborate? For k large, all ${(2k - 1)^2}N$, ${(2k - 3)^2}N$, etc. are smaller than k!, so why do you choose p>k? Of course $p\nmid k!$, because p>k.2012-07-23
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    If $p$ is a prime larger than $k$, then $p\not\mid k!$, by unique factorization. $k!$ cannot have any prime factors larger than $k$2012-07-23
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    Yes, I agree. But are all ${(2k - 1)^2}N$, ${(2k - 3)^2}N$, etc. primes?2012-07-23
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    They are all composite, so that is why I think there may exist at least one k.2012-07-23
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    @Thomas Andrews Never mind. Thanks for help.2012-07-24

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For example, for $R$, you want $k!/(k-j)$ to be a multiple of $N$ for each $j$ from $0$ to $k-1$. That will certainly be true if $k \ge 2N$.

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    Could you explain your reasoning in more detail? How would I apply the same reasoning to $S$ and $T$? Also, I think you mean $k>2N$ because $q$ is an arbitrary integer.2012-07-23