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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?

J__n
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    By simpler multiples $,f\mid x^5!-!1\Rightarrow f(x^5)\bmod f = (f(\color{#0a0}{x^5})\bmod \color{#0a0}{x^5}!-!\color{#c00}1)\bmod f = f(\color{#c00}1)\ \ $ – Bill Dubuque Mar 14 '23 at 18:52
  • If you seek further feedback on your argument then you need to explain how you obtained the 2nd equality in the displayed equation string. Generally, for a solution-verification question to be on topic you must specify precisely which step in the proof you question, and why so (this site is not meant to be a proof checking machine). – Bill Dubuque Mar 14 '23 at 19:03
  • @BillDubuque Thanks for that link to an explanation! I understand how I was mistaken in not being clear with what I'm asking, so, in the future, I will try to be more precise with the step where I need verification. Once again, thank you! – J__n Mar 15 '23 at 13:48

1 Answers1

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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 $f(X)$.

Robert Israel
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    +1 but did you really mean "$X>1$"? In fact I am a bit confused by your first sentence. – peter a g Mar 14 '23 at 18:48
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    I think OP means the polynomial remainder of $f(X^5)$ divided by $f(X)$. In other words, I interpreted $X$ as the indeterminant in the polynomial ring $R[X]$, not a member of $R$. – L. F. Mar 14 '23 at 18:50
  • Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Mar 14 '23 at 18:52