Exercise 11.15

Proof. $\text{∙}$ If $p=13$, the only finite points on the curve are the $4$ points $(0,1)(0,-1),(1,0),(-1,0)$. We must add the $2$ points at infinity to obtain the $6$ points $[t,x,y]$

Since $p=13=3^2+2^2$, where $4\nmid 2$ and $3\equiv -1(\mathit{mod}4)$, here $a=3$, thus the formula of ¤5 gives

$\text{∙}$ If $p=17$, the finite points on the curve, given by the following naive program, are the 12 points

With the two points at infinity, we obtain $14$ projective points.

Here $p=1^2+4^2$, and $p\mid b=4$, $a=1\equiv 1(\mathit{mod}4)$, thus $a=1$, and the formula of ¤5 gives

The formula is verified in both cases.

Program Sage to obtain the finite points on the curve $x^2+y^2+x^2{y}^2=1$:

def N(p):
    Fp = GF(p)
    l = []
    for x in Fp:
        for y in Fp:
            if x^2 + y^2 + x^2*y^2 == 1:
                l.append((x,y))
    return l

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