Proofs Using Results from Cyclotomy

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 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 $f(x) = x^{2} + x + \frac{1} {4}(1 - (-1)^{\frac{p-1} {2} }p) = 0$.