Does $30$ always divide $n^5-n$ where $n \in \mathbb{Z^+}$?

Question

Does $30$ always divide $n^5-n$ where $n \in \mathbb{Z^+}$ ?

Being divisible by $30$ means being divisible by $2,3$ and $5$ simultaneously. Test those separately. — Arthur, Oct 14 '15 at 16:31
Factorisation as $(n^2+1)(n-1)(n+1)n$ shows divisibility by $3$. Fermat's Little Thm shows $n^5\equiv n\pmod{5}$, and hence divisibility by $5$. — Marconius, Oct 14 '15 at 16:32
rIn particular, you only have to test $n=0,1,2,3,4$. Or, to lighten your load further, $-2,-1,0,1,2$. Since $n^5-n$ is an odd function, this actually means you only need to check $0,1,2$. — Thomas Andrews, Oct 14 '15 at 16:33

score 2 · Accepted Answer · answered Oct 14 '15 at 16:32

For a proof, try factoring: $$ n^5 - n = n(n^4 - 1) = n(n^2 + 1) (n-1) (n + 1) = (n-1)(n)(n+1) (n^2 + 1) $$ The product of three successive integers is always divisible by 6, so all you need to show is that either one of the three integers is divisible by 5, or that $n^2 + 1$ is divisible by 5. If none are divisible by $5$, then the middle one must be of the form $n = 5k+2$ or $n = 5k+3$.

Take the first case and look at $$ n^2 + 1 = (5k+2)^2 + 1 = 25k^2 + 20k + 4 + 1 = 25k^2 + 20k + 5 = 5(5k^2 + 4k + 1). $$ That's clearly divisible by 5.

Now.,..work out something similar for the last case, and you're done.

score 1 · Answer 2 · answered Oct 14 '15 at 16:35

The way I like to do it is try to prove and if it doesn't work, try to find a counter example:

Let's try for small $n$: $1^5-1=0$; $2^5-2=30$; $3^5-3=243-3=240$, and these are all divisible by $30$.

For general $n$, write $n=30r+p$, $0\leq p<30$. Then $$n^5-n=(30r+p)^5-(30r+p)=(30r)^5+5(30r)^4p+10(30r)^3p^2+10(30r)^2p^3+5(30r)p^4+p^5-30r-p$$ The only term which is possibly non divisible by 30 is $p^5-p$. This is what we had begun with, but $p<30$. So now you "simply"have to check that $p^5-p$ is divisible by 30 for $0\leq p\leq 29$. The first three cases are done above, and the rest is an exercise for a rainy day.

score 1 · Answer 3 · answered Oct 14 '15 at 16:53

We have $n^5-n=n(n^4-1)=n(n^2+1)(n^2-1)=(n-1)n(n+1)(n^2+1)$, so a product of three consecutive integers always divides $n^5-n$, so it is always divisible by $2$ and by $3$. Now it suffices to show that it's always divisible by $5$. Use induction on $n$. If $n=0$, it's obvious. Assume it's true for $n=k$, so $k^5-k=5m$ ($m\in\mathbb{Z}$). We prove the result for $n=k+1$:

\begin{eqnarray*} (k+1)^5-(k+1)&=&k^5 + 5k^4 + 10k^3 +10k^2 + 5k +1 -k -1 \\ &=& k^5+5k^4+10k^3+10k^2+4k\\ &=& (k^5-k)+5(k^4+2k^3+2k^2+k)\\ &=& 5(m+k^4+2k^3+2k^2+k)\\ \end{eqnarray*}

So $n^5-n$ is always divisible by $5$ too by induction. This completes the proof.

Does $30$ always divide $n^5-n$ where $n \in \mathbb{Z^+}$?

3 Answers3