July 16th, 2017, 03:22 AM  #1 
Newbie Joined: Apr 2017 From: Bhadohi, U.P., India Posts: 23 Thanks: 1  find the number
find the number

July 16th, 2017, 03:27 AM  #2 
Senior Member Joined: Oct 2009 Posts: 275 Thanks: 92 
Write $n^5 + 5 = (n^5  5^5) + (5^5 + 5)$.

July 16th, 2017, 10:38 AM  #3 
Newbie Joined: Apr 2017 From: Bhadohi, U.P., India Posts: 23 Thanks: 1 
please explain further

July 16th, 2017, 12:28 PM  #4 
Senior Member Joined: Oct 2009 Posts: 275 Thanks: 92 
Can you show that $n+5$ always divides $n^5 + 5^5$ by factorizing the latter? Then what does that mean for $n^5 + 5 = (n^5 + 5^5)  (5^5 + 5)$? 
July 16th, 2017, 01:07 PM  #5 
Global Moderator Joined: Dec 2006 Posts: 18,704 Thanks: 1530 
Getting the signs right would help. One can use the factor theorem or the remainder theorem.

July 16th, 2017, 04:55 PM  #6 
Newbie Joined: Jul 2017 From: Asia Posts: 2 Thanks: 1  Last edited by skipjack; July 16th, 2017 at 06:07 PM. 
July 16th, 2017, 06:52 PM  #7 
Senior Member Joined: May 2016 From: USA Posts: 922 Thanks: 368  $\dfrac{n^5 + 5}{n + 5} =$ $n^4  5n^3 + 25n^2  125n + 625  \dfrac{3120}{n + 5}.$ 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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