<s>[INST] Finding the remainder when $p(x^7)$ is divided by $p(x)$, where $p(x)=x^6+x^5+ \cdots + x + 1$? I know there are one or two similar questions to this one but I did not want to look at them since I would like just a hint.

Find the remainder when $p(x^7)$ is divided by $p(x)$, where $p(x)=x^6+x^5+ \cdots + x + 1$?

I wrote $p(x)= \dfrac {x^7-1}{x-1}$, so $p(x^7)=\dfrac {x^{49}-1}{x^7-1}$. We are looking for the unique polynomial $R(x)$ with $deg (R)<6$ such that 
$$\dfrac {x^{49}-1}{x^7-1} =Q(x)\left(\dfrac {x^7-1}{x-1}\right)+R(x)$$
Multiplying both sides by $x^7-1$,
$$x^{49}-1=Q(x)p(x)(x^7-1)+R(x)(x^7-1)$$
I know one way to do it would be to find $R(x)$ at $6$ different points and then set up a system of $6$ equations. I want to plug in values which will make the $Q(x)$ term zero because I can't evaluate $Q(x)$. But the values which will make the $Q(x)$ term zero are the $7^{th}$ roots of unity, which will also make the $R(x)$ term zero.  
 [/INST] Write $p(x^7) = q(x)p(x) +r(x)$ where $deg(r) <7$.  
If we say $1,a_1,...a_6$ are all zeros (which are obviously different) of $x^7=1$ then for each $i$ we get:
$$  p(1) = p(a_i^7) = q(a_i)p(a_i) +r(a_i) = r(a_i)$$
Thus $r$ takes the same value for 6 different values and thus $r(x) \equiv  p(1) =7$. 
</s>