Exercise 8.26

Proof.

(a)

By Proposition 8.1.5, $$\begin{array}{llll}\hfill N(y^2+x^4=1)& =\underset{a+b=1}{\sum }N(y^2=a)N(x^4=b)& \hfill & \\ \hfill & =\underset{a+b=1}{\sum }(1+\rho (a))(1+\chi (b)+{\chi }^2(b)+{\chi }^3(b))& \hfill & \\ \hfill & =\underset{i=0}{\overset{1}{\sum }}\underset{j=0}{\overset{3}{\sum }}\underset{a+b=1}{\sum }{\rho }^i(a){\chi }^j(b)& \hfill & \\ \hfill & =\underset{i=0}{\overset{1}{\sum }}\underset{j=0}{\overset{3}{\sum }}J({\rho }^i,{\chi }^j)& \hfill & \end{array}$$

As $J(𝜀,𝜀)=p$, and $0=J(𝜀,\chi )=J(𝜀,{\chi }^2)=J(𝜀,{\chi }^3)=J(\rho ,𝜀)$, we obtain

$$N(y^2+x^4=1)=p+J(\rho ,\chi )+J(\rho ,{\chi }^2)+J(\rho ,{\chi }^3).$$

As $J(\rho ,{\chi }^3)=J(\bar{\rho },\bar{\chi })=\overline{J(\rho ,\chi )}$, and $J(\rho ,{\chi }^2)=J(\rho ,\rho )=J(\rho ,{\rho }^{-1})=-\rho (-1)=-{(-1)}^{(p-1)∕2}=-1$ (since $p\equiv 1(\mathit{mod}4)$). Moreover, by Exercise 8.7,

$$J(\chi ,\rho )=\chi (-1)J(\chi ,\chi )=\pm J(\chi ,\chi )\in \mathbb{Z}[i]:J(\chi ,\rho )=a+\mathit{bi},(a,b)\in {\mathbb{Z}}^2.$$

Thus $N(y^2+x^4=1)=p+2\mathrm{Re}J(\chi ,\rho )+J(\rho ,\rho )=p-1+2a$

In conclusion,

$$N(y^2+x^4=1)=p-1+2a,\mathrm{where}J(\chi ,\rho )=a+\mathit{bi}.$$

(b)

By Exercise 8.8, $$\begin{array}{llll}\hfill \underset{t\in {𝔽}_p}{\sum }\rho (1-t^4)& =\underset{{\lambda }^4=𝜀}{\sum }J(\rho ,\lambda )& \hfill & \\ \hfill & =J(\rho ,𝜀)+J(\rho ,\chi )+J(\rho ,{\chi }^2)+J(\rho ,{\chi }^3)& \hfill & \\ \hfill & =J(\rho ,\rho )+J(\rho ,\chi )+J(\rho ,\bar{\chi })& \hfill & \\ \hfill & =-1+2\mathrm{Re}J(\chi ,\rho )& \hfill & \\ \hfill & =-1+2a& \hfill & \end{array}$$

So $N(y^2=1-x^4)=p-1+2a=p+\underset{t\in {𝔽}_p}{\sum <!--\; FUNCTION\; APPLICATION\; -->}\rho (1-t^4)$.

(c)

Reducing modulo $p$,and writing $\bar{a}\in {𝔽}_p$ the class of $a$, we obtain : $$\begin{array}{llll}\hfill 2\bar{a}& =1+\underset{t\in {𝔽}_p}{\sum }\rho (1-t^4)& \hfill & \\ \hfill & =1+\underset{t\in {𝔽}_p}{\sum }{(1-t^4)}^{\frac{p-1}{2}}& \hfill & \\ \hfill & =1+\underset{t\in {𝔽}_p}{\sum }\underset{k=0}{\overset{(p-1)∕2}{\sum }}\left(\genfrac{}{}{0.0pt}{}{\frac{p-1}{2}}{k}\right){(-1)}^k{t}^{4k}& \hfill & \\ \hfill & =1+\underset{k=0}{\overset{(p-1)∕2}{\sum }}{(-1)}^k\left(\genfrac{}{}{0.0pt}{}{\frac{p-1}{2}}{k}\right)\underset{t\in {𝔽}_p}{\sum }{t}^{4k}& \hfill & \\ \hfill & =1+\underset{k=1}{\overset{(p-1)∕2}{\sum }}{(-1)}^k\left(\genfrac{}{}{0.0pt}{}{\frac{p-1}{2}}{k}\right)\underset{t\in {𝔽}_p^{\text{∗}}}{\sum }{t}^{4k}.& \hfill & \end{array}$$

Let $S_k=\underset{t\in {𝔽}_p^{\text{∗}}}{\sum <!--\; FUNCTION\; APPLICATION\; -->}{t}^{4k},k>0$, and $g$ a generator of ${𝔽}_p^{\text{∗}}$ : $t=g^i,0\le i\le p-2$.

$$S_k=\underset{i=0}{\overset{p-2}{\sum }}{(g^i)}^{4k}=\underset{i=0}{\overset{p-2}{\sum }}{(g^{4k})}^i.$$

If $g^{4k}\ne 1$, $S_k\equiv \frac{g^{4k(p-1)}-1}{g^{4k}-1}=0$, if not $S_k=p-1=-1$.

For every $k$, $1\le k\le (p-1)∕2$,

$$g^{4k}=1⟺p-1\mid 4k⟺\frac{p-1}{4}\mid k⟺k=\frac{p-1}{4}\mathrm{or}k=\frac{p-1}{2},$$

therefore

$$2a\equiv 1-{(-1)}^{\frac{p-1}{4}}\left(\genfrac{}{}{0.0pt}{}{\frac{p-1}{2}}{\frac{p-1}{4}}\right)-{(-1)}^{\frac{p-1}{2}}\left(\genfrac{}{}{0.0pt}{}{\frac{p-1}{2}}{\frac{p-1}{2}}\right)(\mathit{mod}p),$$

$$2a\equiv -{(-1)}^{\frac{p-1}{4}}\left(\genfrac{}{}{0.0pt}{}{\frac{p-1}{2}}{\frac{p-1}{4}}\right)(\mathit{mod}p).$$

(d)

$\text{∙}$ If $p=13$, I choose the primitive root $g=\bar{2}$, and $\chi$ the character of order 4 defined by $\chi (g)=i$ (the only other character of order 4 is $\bar{\chi }$).

Then $J(\chi ,\rho )=-3+2i,a=-3$.

$$N=p-1+2a=6.$$

$2a=-6,-{(-1)}^{\frac{13-1}{4}}\left(\genfrac{}{}{0.0pt}{}{6}{3}\right)=20$ and $-6\equiv 20(\mathit{mod}13)$.

$\text{∙}$ If $p=17$, $g=3$, $\chi (g)=i$, $J(\chi ,\rho )=-1+4i,a=-1$.

$$N=p-1+2a=14.$$

$2a=-2,-{(-1)}^{\frac{17-1}{4}}\left(\genfrac{}{}{0.0pt}{}{8}{4}\right)=-70\equiv -2(\mathit{mod}17)$.

$\text{∙}$ If $p=29$, $g=3$, $\chi (g)=i$, $J(\chi ,\rho )=5+2i,a=5$.

$$N=p-1+2a=38.$$

$2a=10,-{(-1)}^{\frac{29-1}{4}}\left(\genfrac{}{}{0.0pt}{}{14}{7}\right)=3432\equiv 10(\mathit{mod}29)$ ( $3422=118\times 29$).

Note : By Prop. 8.3.1 (and Ex. 8.7), $p=|J(\chi ,\rho ){\text{|}}^2=|J(\chi ,\chi ){\text{|}}^2$, so $a^2+b^2=p$ (where $p=4m+1$).

As $\left(\genfrac{}{}{0.0pt}{}{2m}{m}\right)=2\left(\genfrac{}{}{0.0pt}{}{2m-1}{m-1}\right)$ is even, and as $p$ is an odd prime $a\equiv \pm \frac{1}{2}\left(\genfrac{}{}{0.0pt}{}{2m}{m}\right)(\mathit{mod}p)$.

Since $p\ge 5$, $p=a^2+b^2$ implies $|a|<\sqrt{p}<p∕2$, thus the least remainder of $\frac{1}{2}\left(\genfrac{}{}{0.0pt}{}{2m}{m}\right)$ is $\pm a$.

Moreover, from Wilson theorem, we obtain a square root of $-1$ in ${𝔽}_p$ ( $p=4m+1$) :

Since ${(\bar{b}{\bar{a}}^{-1})}^2=-\bar{1}$ in ${𝔽}_p$, we obtain $b\equiv (2m)!a(\mathit{mod}p)$. The conclusion is the proposition of Gauss, which gives an explicit formula for the solution of $p=a^2+b^2$ :

Proposition Let $p$a prime of the form $p=4m+1$.

If

then $p=a^2+b^2$. □