Proofs Using Results from Cyclotomy

Oswald Baumgart

Chapter Chapter 5 in The Quadratic Reciprocity Law, 2015, pp 45-62 from Springer

Abstract: Abstract 1. Let ρ be a primitive root of the equation $$\frac{x^{p-1}-1} {x-1} = 0$$ , where p is a positive odd prime, and let g be a primitive root modulo p; then we can order the roots of $$\frac{x^{p-1}-1} {x-1} = 0$$ in the following way: $$\displaystyle{\rho,\rho ^{g^{2} },\rho ^{g^{4} },\ldots,\rho ^{g^{p-3} }\quad \text{and}\quad \rho ^{g},\rho ^{g^{3} },\ldots,\rho ^{g^{p-2} }.}$$ The expressions $$\displaystyle{y_{1} =\rho ^{g} +\rho ^{g^{3} } +\ldots +\rho ^{g^{p-2} },\quad y_{2} =\rho +\rho ^{g^{2} } +\ldots +\rho ^{g^{p-3} }}$$ are called quadratic1 periods quadratic period of the cyclotomic equation $$\frac{x^{p-1}-1} {x-1} = 0$$ . Using their property $$\displaystyle{y_{1} - y_{2} = (\rho ^{-1}-\rho )(\rho ^{-3} -\rho ^{3})\cdots (\rho ^{-p+2} -\rho ^{p-2})}$$ and the relation $$\displaystyle{(x -\rho ^{2})(x -\rho ^{4})\cdots (x -\rho ^{2(p-1)}) = x^{p-1} + x^{p-2} +\ldots +1}$$ we find $$\displaystyle{(y_{1} - y_{2})^{2} = (-1)^{\frac{p-1} {2} }p.}$$ Now y 1 + y 2 = −1, hence we get $$\displaystyle{y_{1}y_{2} = \frac{1 - (-1)^{\frac{p-1} {2} }p} {4}.}$$ Thus the two periods y 1 and y 2 are roots of the quadratic equation $$f(x) = x^{2} + x + \frac{1} {4}(1 - (-1)^{\frac{p-1} {2} }p) = 0$$ .

Keywords: Primitive Root; Quadratic Residue; Residue Modulo; Primitive Root Modulo; Complete Residue System (search for similar items in EconPapers)
Date: 2015
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DOI: 10.1007/978-3-319-16283-6_5

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