<s>[INST] Proof that $(n^7-n^3)(n^5+n^3)+n^{21}-n^{13}$ is a multiple of $3$. I proved that $$(n^7-n^3)(n^5+n^3)+n^{21}-n^{13}$$ is a multiple of $3$ through the use of Little Fermat's theorem but i want to know if there exist other proofs(maybe for induction). How can I demonstrate it?
This my proof:
$$n^3(n^4-1)(n^5+n^3)+n^{13}(n^8-1)$$
Now i know that $$n^4-1\equiv 0\pmod 3 (n\neq 3k)$$ and $$n^8-1\equiv 0\pmod 3 (n\neq 3k).$$
Therefore I proved that $(n^7-n^3)(n^5+n^3)+n^{21}-n^{13}$ is a multiple of $3$.
 [/INST] $$(n^7-n^3)(n^5+n^3)+n^{21}-n^{13}=n^6(n^2-1)(n^2+1)^2+n^{13}(n^8-1)=$$
$$=n^6\left((n^2-1)(n^2+1)^2+n^7(n^2-1)(n^2+1)(n^4+1)\right)=$$
$$=n^6(n^2-1)(n^2+1)\left[n^2+1+n^7(n^4+1)\right]=$$
$$=\color{red}{n(n-1)(n+1)}\cdot n^5(n^2+1)\left[n^2+1+n^7(n^4+1)\right]$$
Can you see why the red part is already a multiple of $\;3\;$ ?
</s>