High degree polynom divisibility The problem is:
Show that $x^{44}+x^{33}+x^{22}+x^{11}+1$ is divisible by $x^4+x^3+x^2+x+1$
I am not sure how to approach this problem so any help would be very appreciated. Thank you in advance.
 A: HINT:
Both are in Geometric Series
$$1+x+x^2+x^3+x^4=\frac{1-x^5}{1-x}$$ and $$1+x^{11}+x^{22}+x^{33}+x^{44}=\frac{1-x^{55}}{1-x^{11}}$$
So, $$\frac{1+x^{11}+x^{22}+x^{33}+x^{44}}{1+x+x^2+x^3+x^4}=\frac{(1-x^{55})(1-x)}{(1-x^{11})(1-x^5)}$$
Now using Prove that $\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$, 
$1-x^{55}$ is divisible by $1-x^5,1-x^{11}$ and 
$(1-x^{11},1-x^5)=1-x$
A: ${\rm mod} \dfrac{(x^5\!-\!1)}{(x\!-\!1)}:\,\ \color{#c00}{x^5\equiv 1}\,\Rightarrow\,x^{11n}\equiv x^n (\color{#c00}{x^5})^{2n}\equiv x^n\Rightarrow\,x^{44}\!+\!x^{33}\!+\cdots+1\equiv x^4\!+\!x^3\!+\cdots+1\equiv 0$
A: It is not hard to see that $x^5-1$ divides $x^{55}-1$. 
Thus $(x-1)(x^4+x^3+x^2+x+1)$ divides $(x^{11}-1)(x^{44}+x^{33}+x^{22}+x^{11}+1)$ and therefore 
$x^4+x^3+x^2+x+1$ divides $(x^{10}+x^9+\cdots +x+1)( x^{44}+x^{33}+x^{22}+x^{11}+1)$.
But $x^4+x^3+x^2+x+1$ and $x^{10}+x^9+\cdots +x+1$ are relatively prime (use the Euclidean Algorithm). 
The result follows. 
A: Try plain polynomial long division. If it gives you trouble, try it with a smaller example like $x^4-2x^3+4x^2-x+7$ divided by $x^2-3$. 
In your original problem, you should find the quotient to be....

 $x^{40}-x^{39}+x^{35}-x^{34}+x^{30}-x^{28}+x^{25}-x^{23}+x^{20}-x^{17}+x^{15}-x^{12}+x^{10}-x^6+x^5-x+1$

