Exercise 6.9

Proof.

$\text{∙}$

p=3. Let $\omega =e^{2\mathit{i\pi }∕3}$. Let $g={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{t=0}^2(t∕3){\omega }^t$ the quadratic Gauss sum. Then $g=\omega -{\omega }^2$.

As $1+\omega +{\omega }^2=0$, $g^2={(\omega -{\omega }^2)}^2={\omega }^2-2{\omega }^3+{\omega }^4={\omega }^2-2+\omega =-3$ :

$$g^2=-3.$$

$\text{∙}$

p=5. Let $\zeta =e^{2\mathit{i\pi }∕5}$.

$g={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{t=0}^4(t∕3){\zeta }^t=\zeta -{\zeta }^2-{\zeta }^3+{\zeta }^4.$

Then $g=\alpha -\beta$, where $\alpha =\zeta +{\zeta }^4,\beta ={\zeta }^2+{\zeta }^3$.

$\alpha +\beta =\zeta +{\zeta }^4+{\zeta }^2+{\zeta }^3=-1$.

$\mathit{\alpha \beta }={\zeta }^3+{\zeta }^4+{\zeta }^6+{\zeta }^7={\zeta }^3+{\zeta }^4+\zeta +{\zeta }^2=-1$

So $\alpha ,\beta$ are the two roots of $x^2+x-1$.

$$\begin{array}{llll}\hfill g^2& ={(\alpha -\beta )}^2& \hfill & \\ \hfill & ={\alpha }^2+{\beta }^2-2\mathit{\alpha \beta }& \hfill & \\ \hfill & ={(\alpha +\beta )}^2-4\mathit{\alpha \beta }& \hfill & \\ \hfill & ={(-1)}^2-4(-1)& \hfill & \\ \hfill & =5.& \hfill & \end{array}$$

Note: here we know explicitely $g$ :

if $p=3$, $g=\omega -{\omega }^2=i\sqrt{3}$.

If $p=5$, $g=\alpha -\beta =(-1+\sqrt{5})∕2-(-1-\sqrt{5})∕2=\sqrt{5}$.