Cyclotomic polynomials for primes: Appendix
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.