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July 16th, 2017, 04:22 AM   #1
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find the number

find the number
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July 16th, 2017, 04:27 AM   #2
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Write $n^5 + 5 = (n^5 - 5^5) + (5^5 + 5)$.
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July 16th, 2017, 11:38 AM   #3
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please explain further
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July 16th, 2017, 01:28 PM   #4
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Can you show that $n+5$ always divides $n^5 + 5^5$ by factorizing the latter?

Then what does that mean for $n^5 + 5 = (n^5 + 5^5) - (5^5 + 5)$?
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July 16th, 2017, 02:07 PM   #5
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Getting the signs right would help. One can use the factor theorem or the remainder theorem.
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July 16th, 2017, 05:55 PM   #6
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Last edited by skipjack; July 16th, 2017 at 07:07 PM.
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July 16th, 2017, 07:52 PM   #7
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Quote:
Originally Posted by Shariq Faraz View Post
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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