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I am trying to find a rational parameterization of the curve (variety) $$x^4+a^2x^2=x^2y^2+(h^2+a^2)y^2$$ If I have done my math correctly, this curve has a singularity of multiplicity 2 at the origin, and the projective extension $$x^4+a^2x^2z^2=x^2y^2+(h^2+a^2)y^2z^2$$ has an isolated singularity at (0:1:0). So, the geometric genus should be $$\frac{(4-1)(4-2)}{2}-2-1=0$$ so the curve should have a rational parameterization. I have found a (non-rational) parameterization by parameterizing the associated equation (hyperbola) found by substituting $$u=x^2$$ $$v=y^2$$ which is: $$(x,y)=(\sqrt{-\frac{a^2-(a^2+h^2))t}{1-t}},\sqrt{-t\frac{a^2-(a^2+h^2)t}{1-t}})$$ but I have not been able to find a substitution that will eliminate the radicals and give me a rational parameterization.

Math Keeps Me Busy
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2 Answers2

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The computation of the genus seems to be incorrect --- the singularity at $(0:0:1)$ only decreases the genus by 1.

In fact, the curve is elliptic. This can be checked by using the Riemann-Hurwitz formula for the map $$ (x:y:z) \mapsto (x^2:y^2:z^2) $$ which is a $4:1$ over the curve $$ X^2 + a^2XZ = XY + (h^2 + a^2)YZ, $$ which is a smooth rational curve (provided $h \ne 0$) and branched over the points $(0:0:1)$, $(0:1:0)$, $(-a^2:0:1)$, and $(1:1:0)$.

Sasha
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  • Thank you. I was afraid that I misunderstood the multiplicity of singular points. I am unfamiliar with the Riemann-Hurwitz formula, but will give it a look. – Math Keeps Me Busy Oct 21 '20 at 23:29
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One thing we can do is reduce the number of radicals we have to solve for. If you say $z = \frac{x}{y}$ then $y = xz$ and our equation becomes:

$x^2 + a^2 = z^2(x^2 + h^2 + a^2)$

by dividing out $x^2$. This is 3 squares equaling 2 squares $(X_1^2 + X_2^2 = Y_1^2 + Y_2^2 + Y_3^2)$. One option for parameterizing this is found here: In ℕ⁺, can the sum of three squares equal the sum of two squares? but that uses a lot of quadratic terms which would make it difficult to harmonize $zX_1 = Y_1$ and $zX_2 = Y_3$.

Fortunately, I found a parameterization of $x_1^2 + x_2^2 + x_3^2 - y_1^2 - y_2^2 - y_3^2 = a$ (see https://artofproblemsolving.com/community/c3046h1150063) which makes it easy to set $a = x_3 = 0$ leaving us with the parameterization: $x_1^2 + x_2^2 - y_1^2 - y_2^2 - y_3^2 = 0$ is parameterized by:

$u = \frac{y_3^2}{4}$

$v = \frac{b}{2}$

$(x_1,x_2,y_1,y_2,y_3) = (u + v + \frac{1}{2},u - v + \frac{1}{2},-u + v + \frac{1}{2},u + v - \frac{1}{2},y_3)$

Now we need to solve $x = x_1$, $a = x_2$, $xz = y_1$, $hz = y_2$, $az = y_3$

3 of these are trivial: $x = x_1$, $az = y_3$, and $hz = y_2$ (h only shows up in one equation so $h = \frac{y_2}{z}$)

We can solve another one using our linear variable b (where $v = \frac{b}{2}$):

$b = \frac{(az)^2 - 2(2a - 1)}{2}$ solves $a = x_2$

This leaves us with $xz = y_1$ which is now:

$\frac{z(az)^2}{2} + z(1-a) = 1-a$ and solving for a yields:

$a = \frac{z - 1 \pm \sqrt{-2z^4 + 2z^3 + z^2 - 2z + 1}}{z^3}$

This is just one radical in one variable as opposed to 2 radicals in 3 variables. Hope that helps!

P.S. An example z value is $\frac{2}{3}$ which I obtained from the following python program:

from math import sqrt

from math import gcd

def isSquare(n):

return (int(sqrt(n)) == sqrt(n))

bound = 1000

for y in range(-bound,bound):

for x in range(-bound,bound):

if(x == y or gcd(y,x) != 1):

    continue

eq = -2*(y**4) + 2*(y**3)*x + (y*x)**2 - 2*y*(x**3) + x**4

if(eq > 0):

    if(isSquare(eq)):

        print(y,x)
ServingSpy
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