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July 10th, 2019, 07:47 AM  #1 
Member Joined: Oct 2013 Posts: 60 Thanks: 6  Find all a such that n^aâˆ’n is divisible by aâ‹…(aâˆ’1) for any integer n.
$$ $$ $$ a \cdot (a1) \ \mid \ n^{a}n \ \ \ \forall \ n \in \mathbb{Z} $$ Find all $a$. $$ $$ 
July 12th, 2019, 08:55 AM  #2 
Senior Member Joined: Dec 2015 From: somewhere Posts: 549 Thanks: 83 
Supposing the expression on the right is a polynomial (degree 2) then $\displaystyle a=2 \;$ , $\displaystyle 2(21) \;  \; n^{2}n =n(n1)$. Since $\displaystyle n(n1)$ is the product of two adjacent numbers, they are divisible by 2. Maybe applying (p divides n^pn for pprime) will solve the problem. Last edited by idontknow; July 12th, 2019 at 09:00 AM. 
July 14th, 2019, 12:54 AM  #3  
Member Joined: Oct 2013 Posts: 60 Thanks: 6 
Yes, \(a=2\) is a solution. You gave a nice proof. Quote:
Nevertheless I think \(a\) has to be prime. $$ a \cdot (a1) \ \mid \ n^{a}n \ = \ a \cdot (a1) \ \mid \ n \cdot (n^{a1}1) \ \ \ \forall \ n \in \mathbb{Z} \ .$$ Conjecture: has only solutions if \(\phi (a)= a1\) with \(\phi ()\) denotes Euler's totient function. This implies \(a\) to be prime. By substitution of \(\phi (a)= a1\) we get: $$ a \cdot \phi (a) \ \mid \ n \cdot (n^{\phi (a)}1) \ \ \ \forall \ n \in \mathbb{Z} \ \ , \ a \in \mathbb{P} \ .$$ Let \(a=3\): $$ 6 \ \mid \ n^{3}n \ = \ 6 \ \mid \ n\cdot(n^{2}1) \ = \ 6 \ \mid \ n \cdot(n1)\cdot (n+1) \ \ \ \forall \ n \in \mathbb{Z} \ .$$ So we have 3 adjacent factors on the right side: \((n1)\cdot n \cdot (n+1)\). One of them is divisible by 3, and one or two of them is/are even. Thus the right side is divisible by \(2\cdot3=6 \ \ \ \forall \ n \in \mathbb{Z} \ \). \(a=3\) is a solution! What further solutions can we find? Last edited by Martin Hopf; July 14th, 2019 at 01:12 AM.  
July 14th, 2019, 03:10 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,823 Thanks: 2160 
$a = 7$ is a solution and $a = 43$ may be a solution. Beyond that, I've no idea.


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aâ‹…aâˆ’1, divisibility, divisible, find, integer, naâˆ’n, number theory, proof 
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