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July 16th, 2017, 03:22 AM   #1
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find the number

find the number
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 July 16th, 2017, 03:27 AM #2 Senior Member   Joined: Oct 2009 Posts: 428 Thanks: 144 Write $n^5 + 5 = (n^5 - 5^5) + (5^5 + 5)$. Thanks from JeffM1 and Shariq Faraz
 July 16th, 2017, 10:38 AM #3 Newbie   Joined: Apr 2017 From: Bhadohi, U.P., India Posts: 26 Thanks: 1 please explain further
 July 16th, 2017, 12:28 PM #4 Senior Member   Joined: Oct 2009 Posts: 428 Thanks: 144 Can you show that $n+5$ always divides $n^5 + 5^5$ by factorizing the latter? Then what does that mean for $n^5 + 5 = (n^5 + 5^5) - (5^5 + 5)$? Thanks from Shariq Faraz
 July 16th, 2017, 01:07 PM #5 Global Moderator   Joined: Dec 2006 Posts: 19,178 Thanks: 1646 Getting the signs right would help. One can use the factor theorem or the remainder theorem. Thanks from Shariq Faraz
 July 16th, 2017, 04:55 PM #6 Newbie   Joined: Jul 2017 From: Asia Posts: 2 Thanks: 1 Thanks from Shariq Faraz Last edited by skipjack; July 16th, 2017 at 06:07 PM.
July 16th, 2017, 06:52 PM   #7
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Quote:
 Originally Posted by Shariq Faraz please explain further
$\dfrac{n^5 + 5}{n + 5} =$

$n^4 - 5n^3 + 25n^2 - 125n + 625 - \dfrac{3120}{n + 5}.$

Simple division.

$n \in \mathbb Z \implies x = n^4 - 5n^3 + 25n^2 - 125n + 625 \in \mathbb Z.$

$n > 3115 \implies n + 5 > 3120 > 0 \implies 1 > \dfrac{3120}{n + 5} > 0 \implies$

$x + 1 > x + \dfrac{3120}{n + 5} > x \implies x + \dfrac{3120}{n + 5} \not \in \mathbb Z$

$\therefore n \le 3115.$

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