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 one year ago
Consider the sets A_n = {1 + k(n!) : 1 <= k <= n}, where the parameter k is assumed to be an integer. Show that any two elements of A_n are relatively prime.
 one year ago
Consider the sets A_n = {1 + k(n!) : 1 <= k <= n}, where the parameter k is assumed to be an integer. Show that any two elements of A_n are relatively prime.

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KingGeorge
 one year ago
Best ResponseYou've already chosen the best response.0Well, take \(a=1+k_1(n!)\) and \(b=1+k_2(n!)\) in \(A_n\). and suppose some prime \(p\) divides both \(a\) and \(b\) (there must be such a prime if they are not coprime). Then \(p(1+k_1(n!))\) implies that \(p\) does not divide \(k_1\) or \(n!\). Likewise, \(p\) does not divide \(k_2\) either. WLOG, assume \(k_1>k_2\). Then \(ab=n!(k_1k_2)\), but since \(pa,b\), \(pab\), and \(p\nmid n!\), it must be that \(pk_1k_2\). Hmmm. I've gotten stuck. Give me a minute.

KingGeorge
 one year ago
Best ResponseYou've already chosen the best response.0I'm sorry, but I've got to go now. If I think of something, I'll come back on later tonight and post what I've got.

KingGeorge
 one year ago
Best ResponseYou've already chosen the best response.0I missed something very obvious. We have that \(pk_1k_2\). However, \(k_2<k_1\le n\), So \(0< k_1k_2< n\). That means that \(pn!\). This is a contradiction, so any two elements of the set are coprime.
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