Show that $x^3+x^2+x+1 \mid x^{4m+3}+x^{4n+2}+x^{4q+1}+x^{4s}$ in $\Bbb Z[x]$ Problem:

Show that $x^3+x^2+x+1 \mid x^{4m+3}+x^{4n+2}+x^{4q+1}+x^{4s}$ in $\Bbb Z[x]$ for all $m,n,q,s$ nonnegative integers

So my algebra sucks.
I'm reviewing some algebra and found this problem. I was able to play with it and get a trivial solution that I'm sure is missing the point of the problem.
My (dumb) solution:

We prove by induction.
For $m=n=q=s=0$ we have that $x^3+x^2+x+1 \mid x^3+x^2+x+1$.
Now assume $x^{4m+3}+x^{4n+2}+x^{4q+1}+x^{4s}=p(x)(x^3+x^2+x+1)$.
Then: $$(p(x)+x^{4m+3}(x-1))(x^3+x^2+x+1)=x^{4(m+1)+3}+x^{4n+2}+x^{4q+1}+x^{4s}$$
$$(p(x)+x^{4n+2}(x-1))(x^3+x^2+x+1)=x^{4m+3}+x^{4(n+1)+2}+x^{4q+1}+x^{4s}$$
$$(p(x)+x^{4q+1}(x-1))(x^3+x^2+x+1)=x^{4m+3}+x^{4n+2}+x^{4(q+1)+1}+x^{4s}$$
$$(p(x)+x^{4s}(x-1))(x^3+x^2+x+1)=x^{4m+3}+x^{4n+2}+x^{4q+1}+x^{4(s+1)}$$

However this is very unenlightening. What is the point of the problem? Is there more enlightening solution?

$i,-1,-i$ are all roots of your polynomial. It is divisible by $(x-i)(x+1)(x+i)=x^3 + x^2 + x + 1.$ I suppose I should add that it  is so divisible in $\mathbb Q [x],$ while Gauss theorem on content says it is then divisible in  $\mathbb Z [x] \; .$
