<s>[INST] Prove that $10|n+3n^3+7n^7+9n^9$ Prove that $10|n+3n^3+7n^7+9n^9$ for every $n\in \mathbb N$
Only what i see that 10=5*2 and both number is free numbers, and if I show that $5|n+3n^3+7n^7+9n^9$ and $2|n+3n^3+7n37+9n^9$ that I prove, since 5 and 2 is free numbers i can use Fermat's little theorem such that  $5|n^5-n$ and $2|n^2-n$ but i can not see that somehow help me, do you have some idea?
 [/INST] Method 1:We can try factoring but ... I don't want to.
Method 2: Euler's theorem.  If $\gcd(n, 10) = 1$ then $n^{\phi(10)}=n^4 \equiv 1\pmod {10}$ So $n + 3n^3 + 7n^7 + 9n^9 \equiv n + 3n^3 + 7n^3 + 9n = 10n + 10n^3 \equiv 0 \mod 10$.
But if $\gcd(n,10) \ne 1$???
Well chinese remainder theorem.
$\mod 2$ we have $0^k \equiv 0$ and $1^k \equiv 1$ so $n^k \equiv n$ so
$$n + 3n^3 + 7n^7 + 9n^9 \equiv n + n^3 + n^7 + n^9 \pmod 2$$ and $a^k \equiv a \pmod 2$ so $$n + n^3 + n^7 + n^9 \equiv 4n \equiv 0 \pmod 2$$
$\mod 5$ we have $\gcd(n,5)$ means $n^{5-1} \equiv 1 \pmod {10}$.
If $\gcd(n,5)= 1$ then $$n+3n^3 + 7n^7 + 9n^9 \equiv n + 3n^3 + 2n^3 + 4n \mod 5\\
 \equiv 5n +5n^3 \equiv 0 \mod 5$$  But if $\gcd(n,5) \ne 1$ the $\gcd(n,5) = 5$ so $n + 3n^3 + 7n^7 + 9n^9 \equiv 0 \mod 5$.
So $2|n+3n^3 + 7n^7 + 9n^9$ and $5|n+3n^3 + 7n^7 + 9n^9$ so $10|n+3n^3 + 7n^7 + 9n^9$.  Always.
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Actually since $\phi(10) = \phi(5) = 4$ we get $n+3n^3 + 7n^7 + 9n^9 \equiv 10n + 10n^3$ when $\gcd(n, 5) \ne 0$.  so the only case to consider is if $\gcd(n, 5) = 5$.
i.e. if $5|n$.
Well if $5|n$ then obviously $5| n+3n^3 + 7n^7 + 9n^9$ and just remains to show $n+3n^3 + 7n^7 + 9n^9$ is even (when $5|n$).
Well, ... that's easy.  If $n$ is even then it is the sum of four even terms.  If $n$ is odd it is the sum of $4$ odd terms.
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