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 November 15th, 2018, 07:06 AM #1 Member   Joined: Oct 2012 Posts: 70 Thanks: 0 Large number divisible Find the largest integer k such that k divides n^55-n for all integer n? Last edited by skipjack; November 16th, 2018 at 05:07 AM.
 November 15th, 2018, 08:32 AM #2 Senior Member   Joined: May 2016 From: USA Posts: 1,251 Thanks: 519 Are you sure that this is the entire question? $\text {Find the largest integer k that divides evenly into } x \text { given that } x = n^{55} - n \text { and } n \in \mathbb Z.$ If $|n| \le 1$, there is no largest integer that divides into x because x will equal 0. Otherwise, the largest integer that can divide evenly into x is |x|. Last edited by JeffM1; November 15th, 2018 at 08:40 AM.
 November 15th, 2018, 10:35 AM #3 Senior Member   Joined: Dec 2015 From: Earth Posts: 327 Thanks: 42 $\displaystyle k=n^{55} -n$ Thanks from topsquark
November 15th, 2018, 10:36 AM   #4
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Quote:
 Originally Posted by fahad nasir Find the largest integer k such that k divides n^55-n for all integer n?
Are you, perhaps, trying to find a function k(n)?

-Dan

November 15th, 2018, 10:50 AM   #5
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Quote:
 Originally Posted by idontknow $\displaystyle k=n^{55} -n$
Actually, that does not work. Consider n = 1. Then

$1^{55} - 1 = 1 - 1 = 0.$

And 0 does not divide evenly into 0.

And $n = -\ 2 \implies 2 > (-\ 2)^{55} - (-\ 2) \text { and } 2 \ | \ (-\ 2)^{55} - (-\ 2).$

Really stupid problem as posted.

Last edited by JeffM1; November 15th, 2018 at 10:57 AM.

 November 16th, 2018, 05:11 AM #6 Global Moderator   Joined: Dec 2006 Posts: 20,105 Thanks: 1907 n^55 - n is divisible by n - 1, n and n + 1, so it's divisble by 2 × 3 = 6. Hence the largest k is at least 6.

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