Find all irreducible factors of the polynomial $x^{15}-1$ over $\Bbb{Q}$.

Clearly, this is a cyclotomic polynomial.

I can find all the irreducible factors but that is very calculative:

\begin{align} \Phi_1(x)&=x-1\\ \Phi_3(x)&=\frac{x^3-1}{\Phi_1(x)}\\ \Phi_5(x)&=\frac{x^5-1}{\Phi_1(x)}\\ \Phi_{15}(x)&=\frac{x^{15}-1}{\Phi_1(x)\Phi_3(x)\Phi_5(x)}. \end{align} The last equation is very calculative. That's why I am looking for any less time consuming method.


  • $\begingroup$ HINT: $ \left ( x^5 \right )^{3} - 1^{3} $, what lovely formula do you know for this difference? $\endgroup$
    – Fede1
    Nov 17 '20 at 19:50
  • $\begingroup$ @Fede1 I tried it alreday. but it is not useful $\endgroup$ Nov 17 '20 at 19:50
  • $\begingroup$ $\Phi_5(x)=\frac{x^5-1}{\Phi_1(x)\Phi_3(x)},$ is wrong $\endgroup$ Nov 17 '20 at 19:56
  • $\begingroup$ @mathcounterexamples.net Yes, Edited $\endgroup$ Nov 17 '20 at 19:58

You have $$\Phi_1(x) \Phi_5(x)= x^5-1$$


$$\Phi_{15}(x)= \frac{x^{15}-1}{(x^5-1) \Phi_3(x)}= \frac{(x^5)^3-1}{(x^5-1)\Phi_3(x)}=\frac{x^{10}+x^5+1}{\Phi_3(x) }$$

One division left...

  • $\begingroup$ LoL...It was not calculating. actually, I am very lazy. $\endgroup$ Nov 17 '20 at 20:10
  • $\begingroup$ I see so! Should be a minute with a pen and some paper... or with Sympy. $\endgroup$ Nov 17 '20 at 20:15
  • $\begingroup$ $\Phi_{15}(x)=x^8 - x^7 + x^5 - x^4 + x^3 - x + 1$. Thanks!(+1) $\endgroup$ Nov 17 '20 at 20:22

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