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How would you show that if $d\mid n$ then $x^d-1\mid x^n-1$ ?

My attempt :

$dq=n$ for some $q$. $$ 1+x+\cdots+x^{d-1}\mid 1+x+\cdots+x^{n-1} \tag 1$$ in fact, $$(1+x^d+x^{2d}+\cdots+x^{(q-1)d=n-d})\cdot(1+x+\cdots+x^{d-1}) = 1+x+x^2 + \cdots + x^{n-1}$$

By multiplying both sides of $(1)$ by $(x-1)$ we get that $1-x^d\mid 1-x^n$ which is the final result

Is this an ok proof?

Community
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John Cataldo
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  • How do you know, or from where do you get, the very first relation to the right of your $;(1);$? – DonAntonio Apr 21 '18 at 16:06
  • @DonAntonio If you divide $x^d-1$ by $x-1$ you should get the series. Its an identity. – CogitoErgoCogitoSum Apr 21 '18 at 16:10
  • @CogitoErgoCogitoSum Oh, I know you must get that...but how? That precisely is part of the proof, as far as I see it... – DonAntonio Apr 21 '18 at 16:10
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    See my edits for proper MathJax usage. In particular, I change $a|b$ to $a\mid b$ and used \cdots and \tag. – Michael Hardy Apr 21 '18 at 16:20
  • I'm with DonAntonio on this. It seems the only way to get this long identities which are not at all self apperant is from actually proofing this result first. Which should be simply a matte of notint $(x^{n-d} +... + 1)(x^d + 1) = x^n - 1$. – fleablood Apr 21 '18 at 16:34
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    I would disagree, guys. Why dont you prove 1+1=2 while youre at it. Its unnecessary. I agree more could have been shown, but I disagree that this needs to be proved or even so much as cited. This is one of those properties that can be found in most any algebra book and would probably be prerequisite knowledge to this particular problem. – CogitoErgoCogitoSum Apr 21 '18 at 16:39

6 Answers6

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An idea for you:

$$d\,\mid\,n\implies n=qd\;,\;\;q\in\Bbb Z\;,\;\;\text{and from here}: $$

$$x^n-1=\left(x^d\right)^q-1=\left(x^d-1\right)\left(\left(x^d\right)^{q-1}+\left(x^d\right)^{q-2}+\ldots+x^d+1\right)$$

The above uses the basic relation from geometric series:

$$x^a-1=(x-1)(x^{a-1}+x^{a-2}+\ldots+x+1)\;,\;\;a\in\Bbb N$$

DonAntonio
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  • Since the question asked about the attempt at a proof, perhaps it would be better to give a hint rather than simply display the answer. –  Apr 21 '18 at 16:11
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    @user43687 Perhaps...yet the OP already wrote down a good try, so giving out the basic stuff isn't bad imo. And the question did not ask about the attempted proof but "How would you show..." , and then it gives it a try. – DonAntonio Apr 21 '18 at 16:12
  • To be honest I was just wondering if my proof is good. Thanks for the answer anyway, I think you wrote q instead of k – John Cataldo Apr 22 '18 at 07:50
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Let $\alpha$ be a solution to the equation $x^d = 1$. We then have that $\alpha^d$ = 1. Since $d|n$ we can write $n = d\cdot k$ for some integer $k$. Thus $$1 = 1^k = \left( \alpha^d \right)^k = \alpha^{d\cdot k} = \alpha^n$$ This shows that $\alpha$ is a solution to $x^n = 1$.

This shows that $x^d-1|x^n-1$.

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You can always do:

$f(y)=1+y+ \dots +y^{r-1}$ so that $yf(y)=y+y^2+\dots +y^r$ and $yf(y)-f(y)=(y-1)f(y)=y^r-1$

Then put $y=x^d$ with $dr=n$ and obtain $(x^d-1)f(x^d)=x^n-1$ and by construction $f(x^d)$ is a polynomial.

Mark Bennet
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Hint from Dummit and Foote:

Suppose $n= kd+r$, where $0 \leq r < d$ we have $$x^n -1 = x^{kd+r} - x^r + x^r -1 = x^r (x^{kd}-1) + (x^r -1)$$ then can you conclude $d|n$ if and only if $x^d-1 | x^n-1$.

Xiao
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Notice that for any $x$ and and natural $n$ that $$(x-1)(x^{n-1} + ..... + x + 1) = (x^n + x^{n-1} +....... +x) - (x^{n-1} + x^{n-2} +.... +1) = x^n -1$$ so that $x-1|x^n - 1$ always.

Lemma: $x-1|x^n-1$ for natural $n$.

Now $d|n$ so let $m = \frac nd$ and let $y= x^d$.

Then $y-1|y^m -1$.

But $y-1 = x^d -1$ and $y^m -1 = x^n - 1$.

In particular: $(x^d -1)(x^{n-d} + x^{n-2d} + ..... + x^d + 1) = x^n - 1$

fleablood
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You have: $$d\mid n \Rightarrow n=ad$$ Then: $$x^n-1=x^{ad}-1=(x^d)^a-1$$ Setting $x^d=y$ (just for simplifying the process) we have: $$y^a-1=(y-1)(y^{a-1}+\dots+y+1)=(x^d-1)((x^d)^{a-1}+\dots + x^d+1)$$ In other words we showed that: $$x^n-1= (x^d-1)((x^d)^{a-1}+\dots + x^d+1) $$ Which obviously implies that: $$x^d-1 \mid x^n-1$$

Anastassis Kapetanakis
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  • You missed the opening brackets at the end of lines 6 and 8. Moreover, this method is identical with that of DonAntonio. – farruhota Apr 21 '18 at 20:33