Exercise 11.14

Proof. We count the number of points at infinity of the curve $C:x^4+y^4=1$ over a finite field $F$. The projective closure of $C$ has equation $x^4+y^4=t^4$. The projective points $[t,x,y]$ such that $t=0$ satisfy the equation $x^4+y^4=0$. Note that $y=0$ is impossible since $[0,0,0]$ is not a projective point. Thus the points at infinity of the curve $C$ are the points $[0,x,y]$ such that $(0,x,y)=y(0,a,1)$, where $a^4=-1$, so that the points at infinity are

Since $(0,a,1)=\lambda (0,b,1)$ for some $\lambda \in F$ implies $a=b$, their number is $N(a^4=-1)$.

Write, as in Chapter 8 and Exercise 8.16, for $a\in F$,

If ${\delta }_4(-1)=0$, then $N(a^4=-1)=0$, and if ${\delta }_4(-1)=1$, then $N(a^4=-1)=4\wedge (p-1)=4$ because $p\equiv 1(\mathit{mod}4)$. In both cases $N(a^4=-1)=4{\delta }_4(-1)$.

To conclude, the number of points at infinity of the curve $C:x^4+y^4=1$ over a finite field $F$ is $4{\delta }_4(-1)$.

In Exercise 8.16, we show that the number of affine points of $C$ is

Therefore the number of points of the projective closure of $C$ in ${\bar{H}}_f(F_p)$ is

With the same calculation as in Exercise 16 and above, we obtain similarly in the field $F_{p^s}$,

where ${\chi }^{\prime }=\chi \circ N_{F_{p^s}∕F_p}$ is a character of order $4$ on $F_{p^s}$.

The generalization of Exercise 8.7 gives

where ${\chi }^{\prime }(-1)=\chi {(-1)}^s={({(-1)}^{\frac{p-1}{4}})}^s$.

As in exercise 13, the Hasse-Davenport relation shows that

Putting all together, we obtain

that is

Then Exercise 2 gives

Using $|\pi {\text{|}}^2=p$, we conclude

Note: By ¤5 (or Ex. 8.18), we know that $a$ is the unique integer such that $p=a^2+b^2$ where $a+\mathit{bi}\equiv 1(\mathit{mod}2+2i)$. With a simpler formulation $p=a^2+b^2$, and $a\equiv 1(\mathit{mod}4)$ if $4\mid b$, $a\equiv -1(\mathit{mod}4)$ if $4\nmid b$. So we can verify these results for small primes $p$. □