Cyclotomic polynomials for primes: Appendix

By Samdney, Mo 30 Juli 2018, in category Math

cyclotomic polynomials, number theory, primes

Ok, last time, we had

\begin{alignat}{3} \phi_{p}\left(x\right) &= \sum_{k = 0}^{p-1} x^{k} \notag \\ &= \sum_{k = 0}^{p-1} e^{k\ln\left(x\right)} \notag \\ &= e^{0\ln\left(x\right)} + \sum_{k = 1}^{p-1} e^{k\ln\left(x\right)} \notag \\ &= 1 + \frac{e^{\ln\left(x\right)} \left(e^{\left(p-1\right)\ln\left(x\right)} - 1\right)}{e^{\ln\left(x\right)} - 1} \end{alignat}

. All what I want to add in this small appendix is, that we can, of course, also write here

\begin{alignat}{3} \phi_{p}\left(x\right) &= 1 + \frac{e^{\ln\left(x\right)} \left(e^{\left(p-1\right)\ln\left(x\right)} - 1\right)}{e^{\ln\left(x\right)} - 1} \notag \\ &= 1 + \frac{x\left(x^{p-1} - 1\right)}{x - 1} \\ &= 1 + \frac{x^{p} - x}{x - 1} \notag \end{alignat}

Now, if we solve this for p

\begin{alignat}{3} \phi_{p}\left(x\right) &= 1 + \frac{x^{p} - x}{x - 1} \notag \\ \phi_{p}\left(x\right) - 1 &= \frac{x^{p} - x}{x - 1} \notag \\ \left(\phi_{p}\left(x\right) - 1\right)\left(x - 1\right) &= x^{p} - x \notag \\ \left(\phi_{p}\left(x\right) - 1\right)\left(x - 1\right) + x &= x^{p} \notag \\ p &= \ln\left(\frac{\left(\phi_{p}\left(x\right) - 1\right)\left(x - 1\right) + x}{x}\right). \end{alignat}

That's it.