When does $120$ divide $n^5 - n$ Find n such that $n^5 - n$ is divisible by $120$. (*Here $n$ belongs to the natural number set)
Seems a trivial problem, but the answer mentioned in the text book seemed incomplete to me so I'm here just to ensure whether approach to this problem is correct or not.
So my approach is as follows:
We know that $n^5 - n$ can be written as $n^5-n = n(n-1)(n+1)(n^2 + 1)$ 
Now since it's product of three consecutive natural numbers so it's divisible by $3! = 6$. Now using Fermat's Little theorem we can prove that $n^5= n \mod5$ hence $n^5 - n$ is divisible by $5$ as well. So $n^5 - n$ is divisible by $30$. Now to ensure their is another factor of $4$ in the term we need to ensure that $n^5 - n$ is divisible by $8$. I found that all odd natural numbers and multiples of $8$ makes $n^5 - n$ divisible by $8$, hence $n^5 - n$ is divisible by $120$. But in the text book they have only mentioned odd natural numbers as answer, which seemed incomplete to me.
Thanking everyone in advance:P    
 A: You're right and the textbook is wrong.
For odd $n$, $n-1$ and $n+1$ are even, and one of them is divisible by $4$; hence the product is divisible by $8$.
For even $n$, neither $n-1$, nor $n+1$, nor $n^2+1$ is even, so factors of two only come from $n$. Thus, for even $n$, $n^5-n$ is divisible by $8$ (and thus by $120$) if and only if $n$ itself is divisible by $8$.
A: We need $8$ to divide $$n(n^4-1)$$
As $n(n^3)+(-1)(n^4-1)=1,(n,n^4-1)=1$
So, either $8|n$
Or $n$ must be odd and 
$$8|(n^4-1)\iff2\mid\dfrac{n^2-1}2\cdot\dfrac{n^2+1}2$$ which is true for all odd $n$ as the later is the product of two consecutive integers
A: You are correct. It's case $p^k = 2^3,\,$ and $\, f = \phi(p^k) = 4\,$ below
Lemma $ $ If $\,p\,$ is prime and $\,\phi(p^k)\mid f>0\,$ then $\,p^k\mid n(n^f-1)\iff p^k\mid n\,$ or $\,(p,n)=1$
Proof $\,\ (n,\,n^f-1) = 1\,$ so $\,p^k\mid n(n^f-1)\iff p^k\mid n\,$ or $\,p^k\mid n^f-1,\,$ and the latter holds iff $\,(p,n)=1,\,$ since then $\,\phi\mid f,\, n^\phi\equiv 1\Rightarrow\, n^f\equiv 1\pmod{\!p^k}\,$ by Euler & modular order reduction.
Remark $ $ Stronger, we can replace Euler phi by Carmichael lambda (= half of phi for $2^k,\ k>2)$ 
See also this generalization of Euler and Fermat.
