Show that for every $n \in \mathbb{N}$ $120 \mid n^7 - n^3$

Question

I want to show that for every $n \in \mathbb{N}: 120 \,\mid\, n^7 -n^3$.

Currently I only have:

$$ 120 \,\mid\, n^7 - n^3 \\ \Leftrightarrow n^7 - n^3 \equiv 0 \,\,\,\,\text{(mod 120)} \\ \Leftrightarrow n^7 \equiv n^3 \,\,\,\,\text{(mod 120)} $$

How can I proceed from here?

Answer 1

We want $120=8\cdot 3\cdot 5$ to divide $n^3(n^4-1)=n^3(n-1)(n+1)(n^2+1)$.

8, 3, 5 are relatively prime, so we can consider them independly.

If $n$ is even, $8|n^3$; if $n$ is odd then $2|n^2+1$ and $n\pm 1$ both even and one of them divisible by 4, so the product is divisible by 8. This gives $8|n^7-n^3$.

If $3|n$, then $3|n^7-n^3$ as well; if $3\nmid n$, then $3|n^2-1$ by Fermat's theorem.

If $5|n$, then $5|n^7-n^3$ as well; if $5\nmid n$, then $5|n^4-1$ by Fermat's theorem.

Answer 2

$120 = 2^3\cdot 3\cdot 5$

Show that $n^7 = n^3 \pmod {3,5,8}$

Euler / Fermat will help

Or you can factor it.

$n^7 - n^3 = (n^3)(n^2 + 1)(n+1)(n-1)$

And show how the factors must be divisible by $3,5, 8$

Or you can prove by induction.

Answer 3

Show that $n^7\equiv n^3$ modulo $3,5$ and $8$.

Modulo $3$, use the "extended" version of Fermat's Little Theorem: if $p$ is prime then $n^p\equiv n\pmod p$ for all $n$. So we have $$\eqalign{n^3\equiv n\quad &\Rightarrow\quad n^7\equiv n^5\ \hbox{and}\ n^5\equiv n^3\cr &\Rightarrow\quad n^7\equiv n^3\ .\cr}$$

Modulo $5$, very similar (in fact a bit easier).

Modulo $8$, if $n$ is odd then $n^2\equiv1$ so $$n^7\equiv n^5\equiv n^3\ .$$ If $n$ is even please try it yourself.