Let $f=X^4+X^3+X^2+X+1$. Determine the remainder of dividing $f(X^5)$ to $f(X)$.
My question is whether my thinking is correct in solving this problem, as I want to make sure I can apply the same logic for other problems like this one.
What I noticed is that $X^4+X^3+X^2+X+1 = 0 \iff X^5 = 1$. Then, I know that the remainder of $g : (X-\alpha)$ is equal to $g(\alpha)$, for $g$ an element of a field. Therefore: $$\text{remainder-of}\left(\frac{f(X^5)}{f(X)}\right) = \text{remainder-of}\left(\frac{f(X^5)}{X^4+X^3+X^2+X+1}\right) = \\[4ex] =\text{remainder-of}\left(\frac{\overbrace{X\cdot f(X^5)}^{g(X)}}{\underbrace{X^5-1}_{X-1}}\right)=g(1)=1\cdot f(1)=1\cdot(1+1+1+1+1)=5$$
So, is this correct/rigorous? Can the same logic be used for similar tasks, or it just so happens that this process works for this one problem?
The remainder is $5$ for $X > 1$, but not for $X=0$ or $X=1$. But it's true that $f(X^5) \equiv 5 \mod f(X)$.
The way I'd say it is that $Y^{j}-1$ is divisible by $Y-1$ for natural numbers $j$, so in particular $X^{5j}-1$ is divisible by $X^5-1$. Then $$f(X^5) - 5 = (X^{4\cdot 5} - 1) + (X^{3\cdot 5}-1) + (X^{2\cdot 5}-1) + (X^{1\cdot 5}-1) + (X^{0\cdot 5}-1)$$ is divisible by $X^5-1$, and therefore by $f(X)$.