<s>[INST] What is the largest number that divides $n^5-n$ whenever $n$ is odd? My attempt: Let $n=2k+1$, where $k$ is an Integer. Then $$n^5-n=(2k+1)(2k)2(k+1)2(2k^2+2k+1)=8(2k+1)(k)(k+1)(2k^2+2k+1)=16(2k+1)(P)(2k^2+2k+1)$$ 
where $P$ is an integer. So according to me the answer should be atleast $16$. How should I go about finding other factors from this expression. What other approaches can I use to solve problems of this kind? 
Edit: This was a multiple choice question. The choices were:


*

*15

*30

*16

*240

*720

 [/INST] With such problems, you should always factorize first and see if you can get anywhere from there. Also, plugging in specific values helps (do that with the factorized expressions). That already suffices here:
Okay, so let $N = n^5 - n = n·(n^4 - 1) = n·(n-1)·(n+1)·(n^2 + 1)$ for some $n ∈ ℕ$.
If $n$ is odd,


*

*all of $n-1$, $n+1$ and $n^2 + 1$ are even, one of $n-1$, $n+1$ even being divisible by $4$, so $N$ is divisible by $16$,

*one of $n$, $n-1$ and $n+1$ is divisible by $3$, so $N$ is divisible by $3$,

*either one of $n$, $n-1$ or $n+1$ is divisible by $5$ or else $n \equiv \pm 2 \bmod 5$, so $n^2 + 1 \equiv 5 \equiv 0 \bmod 5$, and $N$ is divisible by $5$.


So all numbers $N = n^5 - n$ with $n$ being odd are divisible by $16·3·5 = 240$.
For $n = 3$, $N = 3·2·4·10 = 240$, so there is no number larger than $240$ that divides all numbers $N = n^5 - n$ with $n$ being odd.
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