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Theorem: If $a$ is a special root of the equation $x^{n}-1=0$, then $a^{p}$ is also a special root of it (where $p$ is prime to $n$).

I have done a proof of this theorem. Can you please tell if my method is incorrect?

My Method: as $a$ is a special root of it, then we can express $a$ as $$ \cos\left(\frac{2r\pi}{n}\right) + i\sin \left(\frac{2r\pi}{n}\right) $$ where $r$ is prime to $n$.

Now as $p$ is also prime to $n$, then $\gcd(pr,n) = 1$ so $$ a^{p} = \cos\left(\frac{2pr\pi}{n}\right) + i\sin \left(\frac{2pr\pi}{n}\right) $$ is also a special root of it. (Proved)

This was my method. Thank you very much. Please tell me if there was a mistake. I do not really know the math programming inputs. So, I'm so sorry if it is hard for you to understand what I've written.

egreg
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Avijit Mukherjee
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1 Answers1

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$(a^p)^n=(a^n)^p=(1)^p=1$

Let $a^p$ is a root of $x^r-1=0$ where $1\le r<n$

$\implies(a^p)^r=1\iff a^{pr}=1$

as $a$ is a special root of $x^n-1=0, pr\not<n$

As $(p,n)=1, pr>n$

Let $pr=nq+R$ where $0\le R<n$

$a^R=a^{pr-nq}= \dfrac{a^{pr}}{(a^n)^q}=\dfrac11$

$\implies R=0\implies pr=nq\iff\dfrac{pr}n=q$ which is an integer

$\implies n|pr$

$(n,p)=1\implies n|r$ but $r<n\implies r=0$ which is a contradiction

lab bhattacharjee
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  • thank you very much @lab bhattacharjee ..can you pls tell me what is wrong with my method.....?And do I need to tell that r lies between 1 to n? because any integer that is prime to n can be used to express the special root – Avijit Mukherjee Aug 23 '15 at 14:01