show that $11\nmid n^6-2n^5+4n^4-8n^3+16n^2-32n+64 $ 
Let $n$ be postive integer,show that
  $$11\nmid n^6-2n^5+4n^4-8n^3+16n^2-32n+64 $$

I have use ugly metods to solve consider this case,$n\equiv 0,\pm 1,\pm 2,\pm 3,\pm 4,\pm 5\pmod {11}$,have other more simple methods it?
 A: Starting from @Crostul's comment, we have
$$
f(n)=n^6-2n^5+4n^4-8n^3+16n^2-32n+64 = \frac{n^7+2^7}{n+2}
$$
and so $f(n)(n+2)=n^7+2^7$.
Therefore, if $11$ divides $f(n)$, then $11$ divides $n^7+2^7$.
Now $n \mapsto n^7$ is an injective map mod $11$ because $\gcd(7,\phi(11))=1$.
Therefore, if $11$ divides $f(n)$, then $n^7 \equiv -2^7 = (-2)^7 \bmod 11$ and so $n \equiv -2 \bmod 11$.
So you only have to check that $11$ does not divide $f(n)$ for $n=-2$.
A: Suppose $11\mid n^6-2n^5+4n^4-8n^3+16n^2-32n+64$. Then $n^7 + 2^7 = 0 (\mod 11)$ therefore $n^7 + 7 = 0 (\mod 11)$ equivalent $n^7= 4 (\mod 11)$. Using Fermat's little theorem, we get $1 = 4n^3(\mod 11) $ equivalent $3 = n^3(\mod 11)$
From $n^7= 4 (\mod 11)$ we get $9n = 4 (\mod 11) $ therefore $n=9(\mod 11) $. Now  $n^6-2n^5+4n^4-8n^3+16n^2-32n+64 (\mod 11)$ for $n=-2$ becomes $7 \cdot2^6 (\mod 11)$ which is $\ne 0 \mod 11$
A: Let 
$$P(n)=n^6-2n^5+4n^4-8n^3+16n^2-32n+64$$  
Then $P(n+11)\equiv P(n)$ mod $11$, so if $11\mid P(n)$ for some integer $n$, then $11\mid P(n)$ for some even integer $n=2m$.  But $P(2m)=64(m^6-m^5+m^4-m^3+m^2-m+1)$.  If $11\mid P(2m)$, then, since $64\not\equiv0$ mod $11$, we must have
$$m^6-m^5+m^4-m^3+m^2-m+1\equiv0\mod11$$
After checking that $m=-1$ is not a solution, since 
$$(-1)^6-(-1)^5+(-1)^4-(-1)^3+(-1)^2-(-1)+1=7\not\equiv0\mod11$$
we can multiply both sides by $m+1$ to get the equivalent condition
$$m^7+1\equiv0\mod11$$
or $m^7\equiv-1$ mod $11$.  Cubing both sides implies $m^{21}\equiv-1$ mod $11$.  After checking that $m=0$ is not a solution, we note that $m^{10}\equiv1$, which leads to $m\equiv-1$ mod $11$, which has already been ruled out.
A: Guess $11|n^6−2n^5+4n^4−8n^3+16n^2−32n+64$, then $11|n^7+2^7$ so $n^7 \equiv-2^7 \pmod {11}.$ By Euler's theorem, $n^{10} \equiv 1 \pmod {11}$, so $n \Rightarrow n^{21} \Rightarrow (n^7) ^3 \Rightarrow -2^{21} \Rightarrow \equiv -2 \pmod {11} .$ By distributing the exponent, we have $11v(n^7+2^7) \Rightarrow 11v(n+2)+11v(7) \Rightarrow 11(n+2)$, so $(n^6−2n^5+4n^4−8n^3+16n^2−32n+64) = 11v(n^7+2^7)-11v(n+2) = 0$. 
