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Prove that for all $n\in\mathbb{Z}$, $15$ divides $n^7+2n^5+4n^3+8n$.

I know that I have to show that 5 divides $n^7+2n^5+4n^3+8n$ and that 3 divides $n^7+2n^5+4n^3+8n$, I also know that by FLT $n^5 \equiv n \pmod{5}$ and that $n^3 \equiv n \pmod{3}$, but I'm not sure on how to further proceed with this. Any help would be appreciated.

Thanks For the Help guys, I figured it out on my own though!

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Note by Fermat's little theorem, for any non-negative integer $k$, $n^{5+k}\equiv n^{1+k}$ modulo 5. Hence $$n^7+2n^5+4n^3+8n\equiv n^3+2n-n^3+3n\equiv 5n\equiv 0\pmod{5}.$$ A similar approach can be done modulo 3 (or modulo any prime $p$). Can you take it from here?

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Hint $$n^7+2n^5+4n^3+8n=n^4(n^3-n)+3n^2(n^3-n)+7(n^3-n)+15n$$

$$n^7+2n^5+4n^3+8n=n^2(n^5-n)+2(n^5-n)+5n^3+10n$$

Now use for prime $p, n^p\equiv n\pmod p$

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$$n^3-n=n(n^2-1), n^5-n=n(n^2-1)(n^2+1)\implies n^5-n$$ is divisible by $(3,5)=15$

Now $$n^7+2n^5+4n^3+8n=n^5(n^2+2)+4n^3+8n\equiv 5n^3+10n\pmod{15}$$

We need $15|(5n^3+10n)\iff3|(n^3+2n)$

But $n^3+2n=n^3-n+3n$

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The method of finite differences gives for $\displaystyle\Pi_k(n)=\prod\limits_{i=0}^{k-1}(n-i)=k!\binom{n}{k}$

Start calculating $f(n)-\Pi_7(n)=21n^6+...\quad$ the term in $n^7$ has cancelled, and a polynomial of degree $6$ remains.

Then continue by substracting $21\,\Pi_6(n)$ and so on until getting $0$.


$\displaystyle n^7+2n^5+4n^3+8n$

$\displaystyle=\Pi_7(n)+21\,\Pi_6(n)+142\,\Pi_5(n)+370\,\Pi_4(n)+355\,\Pi_3(n)+105\,\Pi_2(n)+15\,\Pi_1(n)$

$\displaystyle=\underbrace{5040}_{15\times 336}\binom{n}{7}+\underbrace{15120}_{15\times 1008}\binom{n}{6}+\underbrace{17040}_{15\times 1136}\binom{n}{5}+\underbrace{8880}_{15\times 592}\binom{n}{4}+\underbrace{2130}_{15\times 142}\binom{n}{3}+\underbrace{210}_{15\times 14}\binom{n}{2}+\underbrace{15}_{15\times 1}\binom{n}{1}$


Binomial coefficients are integers and the numeric ones are all divisible by $15$, so the whole thing is divisible by $15$.

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