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If $n$ is a composite number, I want to prove that $x^{n-1} + x^{n-1} + \ldots + x + 1$ is reducible. I am familar with the factorization \begin{align*} x^n - 1 = (x - 1)(x^{n-1} + x^{n-2} + \ldots + 1). \end{align*} If $n$ is composite, I know I can write $n = xy$ non-trivially. I do not know, however, how to proceed, because the only way I can isolate the polynomial I'm interested in is to redivide by $x - 1$, which is not helpful. I don't think polynomial long-divison will work well either.

Bill Dubuque
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Ted Baker
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  • I fixed it, thank you. Any help with proving this would be appreciated. – Ted Baker Dec 03 '20 at 07:15
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    Hint: If $n = ab$, then $x^n - 1 = (x^b)^a - 1$. Try your factorisation on $u^a - 1$ where $u = x^b$. – user854214 Dec 03 '20 at 07:15
  • Thank you. I will try this and update my answer with another attempt. – Ted Baker Dec 03 '20 at 07:16
  • You picked terrible notation in $n = xy$ when $x$ is your polynomial variable. Write $n = ab$. Tip: Look at $x^n - 1 = x^{ab} - 1$ as $(x^a)^b - 1$. For $b> 1$ you can factor $y^b - 1$ (no typo there with $y$) and for $a > 1$ you can factor $x^a - 1$. Put that information together. – KCd Dec 03 '20 at 07:16
  • By using your suggestion with $n = ab$ and expanding out $(x^a)^b$, I get $x^{n-1} = (x^a - 1)(x^{a(b-1)} + x^{a(b-2)} + \ldots + 1)$. How do I factor $x^a - 1$, though? Do I just plug it write back into the original formula with $a$ in place of $n$? Do I then divide by everything on the right-hand side other than $x^{n-1} + x^{n-2}+ \ldots + 1$? – Ted Baker Dec 03 '20 at 07:22
  • @TedBaker Yes, that's right. Jose did this in his answer. I think you're just missing the step where you also factorise $x^n - 1$ on the left hand side, and cancel the $x - 1$ factors. – user854214 Dec 03 '20 at 07:48
  • Factor theorem $\Rightarrow x^a-1\mid f(x^a)-f(1)\ [,=, x^{ab}-1,$ for $,f = x^a,]\ \ $ – Bill Dubuque Dec 03 '20 at 08:08

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