<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] I will prove the induction method here for $n \geq 0$
Base case : $n=0,1$
Clearly $0$ works.
now $(1^7-1^3)(1^5 + 1^3) + 1^{21}  - 1^13 = 0$ which is also a multiple of $3$
Now assume it works for a value $k \geq 0$ that is $$(k^7-k^3)(k^5 + k^3) + k^{21} - k^{13}$$ is a mulitple of $3$ 
That means that $3m = (k^7-k^3)(k^5 + k^3) + k^{21} - k^{13} = k^{21} -k^{13} + k^{12} + k^{10} -k^{8} -k^{6} $ for some integer $m$
Now we just need to show that this holds for $(k+1)$. That is we need to show that $(k+1)^{21} -(k+1)^{13} + (k+1)^{12} + (k+1)^{10} -(k+1)^{8} -(k+1)^{6}$  is a multiple of $3$
Can you continue from here ??
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