<s>[INST] Find the remainder when $f(x) = \sum_{i=0}^5 x^i$ is divided into $f(x^{12})$ I am looking for confirmation that this proof works. Feel free to be very pedantic.
We wish to find the remainder when $f(x) = \sum_{i=0}^5 x^i$ is divided into $f(x^{12})$. Notice that $f(x)(x - 1) = x^6 - 1$, so let $r$ be a root of $f$. Then $f(r)(r - 1) = 0$ which implies that $r^6 - 1 = 0$, and so $r^6 = 1$. We rewrite the polynomial $f(x^{12})$ into quotient + remainder form as $f(x^{12}) = q(x)f(x) + g(x)$, and so $f(r^{12}) = q(r)f(r) + g(r) = q(r)\cdot0 + g(r) = g(r)$. However, note that $r^{12} = (r^6)^2 = 1$, and so $f(r^{12}) = f(1)$. Now we evaluate $f(1) = \sum_{i=0}^5 1 = 6$, and so the remainder is $6$.
 [/INST] You are almost there. Let us write $Q(x) := f(x^{12})$. Then indeed: $$Q(x) = f(x^{12})=f(x)q(x)+r(x),$$ where $r$ is a polynomial of degree at most $4$.
You noted that $Q(a)=6$ for all $a$ satisfying $a^6=1$, and that the roots of $f$ are $\{a; a^6=1$; $a \not = 1\}$, so that $Q(a)=6$ for each $a$ such that $f(a)=0$, or equivalently, $Q(a)=0$ for each of the roots $a$ of $f$. That is great, but you actually needed to explicitly note is that the equation $Q(a)=6$ holds for each of the $5 >4 =$deg$(r)$ roots $a$ of $f$. And thus, as $r(a)=6$ for $5 >$ deg$(r)$ values of $a$, it follows that $r$ has to be $6$ everywhere, as the only such polynomial $r$ of degree $4$ or less that satisfies $r(a)$ for $5$ or more values of $a$, is $r(x)=6 \ \forall x$ .
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