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ThePerfectHacker · #5 April 11th, 2008, 11:45 AM

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1a)Let $n$ be an odd positive integer. Let $\zeta = e^{2\pi i/n}$ prove that $x^n - y^n = \prod_{k=0}^{(n-1)} (x\zeta^k - y\zeta^{-k})$ .

Note that $z^n - 1 = \prod_{k=0}^{n-1} (z - \zeta^k)$ . Let $z=x/y$ and we get $x^n - y^n = \prod_{k=0}^{n-1} (x-y\zeta^k)$ . Since $0,1,...,n-1$ and $0,-2,-4,...,-2(n-1)$ are both a complete system of residues, since $n$ is odd it means, $\prod_{k=0}^{n-1} (x - \zeta^k y) = \prod_{k=0}^{n-1} (x-\zeta^{-2k}y) = \zeta^{-0-1-...-(n-1)}\prod_{k=0}^{n-1} (x\zeta^k - y\zeta^{-k})$ . But $\zeta^{-0-1-...-(n-1)} = 1$ the proof is complete.

Quote:

1b)Let $f(z) = 2i\sin (2\pi z)$ and $n$ an odd positive integer. Prove that $f(nz) = f(z)\prod_{k=1}^{(n-1)/2} f\left( z + \frac{k}{n} \right) f\left(z - \frac{k}{n} \right)$ .

This function has the property that $f(-z) = -f(z)$ and $f(z+1)=f(z)$ . Also, $f(z) = e^{2\pi i z} - e^{-2\pi i z}$ .
In the identity, $x^n -y^n = \prod_{k=0}^n (x\zeta^k - y\zeta^{-k})$ let $x=e^{2\pi i z}$ and $y=e^{-2\pi i z}$ . Thus, we get, $f(nz) = e^{2\pi i n z} - e^{-2\pi i n z} = \prod_{k=0}^{n-1} \left( e^{2\pi i z} \zeta^k - e^{-2\pi i z} \zeta^{-k} \right)$ . Note that, $e^{2\pi i z} \zeta^k - e^{-2\pi i z} \zeta^{-k}= e^{2\pi i z} e^{2\pi i k/n} - e^{-2\pi i z} e^{-2\pi i k/n} = e^{2\pi i \left( z + \frac{k}{n} \right)} + e^{-2\pi i \left( z + \frac{k}{n} \right)} = f\left( z + \frac{k}{n}\right)$ . This means, $f(nz) = \prod_{k=0}^{n-1}f\left( z + \frac{k}{n} \right)$ . Split the product to get, $f(nz) = f(z) \prod_{k=1}^{(n-1)/2} f\left( z + \frac{k}{n} \right) \prod_{k=(n-1)/2}^{n-1} f\left( z + \frac{k}{n} \right)$ . We know that $f(z) = f(z-1)$ (a property mentioned above) thus $f\left( z + \frac{k}{n} \right) = f\left( z + \frac{k}{n} - 1\right) = f\left( z - \frac{n-k}{n} \right)$ . Thus, $f(nz) = f(z) \prod_{k=0}^{(n-1)/2} f\left( z + \frac{k}{n} \right) \prod_{k=(n-1)/2}^{n-1} f\left( z - \frac{n-k}{k} \right)$ . But $\prod_{k=(n-1)/2}^{n-1} f\left( z - \frac{n-k}{k} \right) = \prod_{k=1}^{(n-1)/2} f\left( z - \frac{k}{n}\right)$ because the products run through the same values. And we have proven that $f(nz) = f(z) \prod_{k=1}^{(n-1)/2} f\left( z + \frac{k}{n} \right) f\left( z - \frac{k}{n}\right)$ .
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This identity belongs to Leopold Eisenstein. With it we can give a very nice short proof of the Quadradic Reciprocity Law. In fact, Enrst Kummer called this (Eisenstein's) proof to be the most beautiful of all reciprocity proofs (Taken from my Number Theory book).

If anybody wants I can post Eisenstein's proof.

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