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--- finite_geometric_series [2016/06/10 01:35]
nikolaj
+++ finite_geometric_series [2016/06/10 01:41]
nikolaj
@@ Line 31: / Line 31: @@
 The proof of the infinitude of primes using Fermat numbers uses this.
->Factorization of $a^n-b^n$:
+In $\mathbb C$, the equation $(x/b)^n=1$ is solved by $x=b\cdot{\mathrm e}^{2\pi i\frac{k}{n}$, so
->
->https://en.wikipedia.org/wiki/Factorization#Sum.2Fdifference_of_two_cubes
+$a^n-b^n=\prod_{k=1}^n (a-b\cdot{\mathrm e}^{2\pi i\frac{k}{n}})=(a-b)\prod_{k=1}^{n-1} (a-b\cdot{\mathrm e}^{2\pi i\frac{k}{n}})$
->
->Is the following true? How to prove it?
->
->$a^n-b^n=\prod_{k=1}^n (a-b\cdot{\mathrm e}^{2\pi i\frac{k}{n}})=(a-b)\prod_{k=1}^{n-1} (a-b\cdot{\mathrm e}^{2\pi i\frac{k}{n}})$
->
 <code>
 n = 6;