Birational transformation between some Diophantine equations and elliptic curves

The cubic equation is birationally equivalent to the elliptic curve E,

E : y² = x³ - 423.A²    (A ∈Z; X, Y ∈Q)

under the following birational transformation,

X = (36A-y)/6x
Y = (36A+y)/6x

and

x = 12A/(X+Y)
y = 36A(Y-X)/(X+Y).

More generally, the curve,

au³ + bv³ + cw³ = 0

is equivalent to the following curve,

r³ + s³ + abct³ = 0

by the following transformation,

r = - 6bc²v³w⁶ - c³w⁹ - 3b²cv⁶w³ + b³v⁹
s = - 3bc²v³w⁶ + c³w⁹ - 6b²cv⁶w³ - b³v⁹
t = - 3uvw (b²v⁶ + bcv³w³ + c²w⁶)

and it is equivalent to

y² = x³ - 432a²b²c²

by the following transformation,

x = 4 (b²v⁶ + bcv³w³ + c²w⁶) / u²v²w²
y = 4 (2b³v⁹ + 3b²cv⁶w³ - 3bc²v³w⁶ - 2c³w⁹) / u³v³w³

This curve is birationally equivalent to,

E : y² = x³ - 4Dx

under the following birational transformation,

x = 2(Y+1) / X²
y = 4(Y+1) / X³

and

X = 2x / y
Y = - 1 + 2x³/y².

More generally, the curve,

v² = au⁴ + bu³ + cu² + du + q²

is birationally equivalent to,

y² + a₁x + a₃y = x³ + a₂x² + a₄x + a₆

under the following birational transformation,

x = (2q(v+q)+du) / u²
y = (4q²(v+q)+2q(du+cu²)-d²u²/2q) / u³

and

a₁ = d/q
a₂ = c - d² / 4q²
a₃ = 2qb
a₄ = -4q²a
a₆ = a₂a₄ = a(d²-4q²c)

The definition of congruent number and the birational transformation between the defined curve and the elliptic curve

E : y² = x(x² - D²)

is described here. The criteria for the congruent number was given by

J. B. Tunell, "A Classical Diophantine Problem and Modular Forms of Weight 3/2", Inventiones mathematicae, 72(1983)323-334.

If there exists the solution of following two quadratic forms simultaneously,

x² + y² = z²
x² + ny² = w²

n is called as concordant. If n is concordant, the following elliptic curve,

E : y² = x(x+1)(x+n)

has a rank more than 0 (i.e. it has non-trivial rational points).

When (u,v) is the point on

v²=u³+(n+1)u²+nu=u(u+1)(u+n)

let a and b

a=v/(n+u)
b=(n+u)(2un+n+u²)/(n+u)²

when

a=p/q

compute x and y by p and q such as

x=p²-q²
y=2pq

then x²+y² and x²+n*y² are perfect square.
Let z and w by

z=sqrt(x²+y²)
w=sqrt(x²+n*y²)

then x, y, z and w satisfy

x² + y² = z²
x² + ny² = w²

Melvyn J. Knight has asked which integers n can be represented as
n=(x+y+z)(1/x+1/y+1/z) where x, y, z are integers.
Rewrite this as,

(x+y+z)(yz+zx+xy)=nxyz

x²(y+z)+x(y²+(3-n)yz+z²)+yz(y+z)=0

x={-y²+(n-3)yz-z²-Δ}/2(y+z)

where

Δ²=y⁴-2(n-1)y³z+(n²-6n+3)y-2z-2-2(n-1)yz³+z⁴

This is birationally equivalent to the elliptic curve

τ²=σ(σ²+(n²-6n-3)σ+16n)
   σ=-4(yz+zx+xy)/z²
   τ=2(σ-4n)y/z-(n-1)σ

and

x,y/z = {±τ-(n-1)σ}/2(4n-σ)

This birational transformation was given by Andrew Bremner, Richard K. Guy, Richard J. Nowakowski,
"Which integers are representable as the product of the sum of three integers with the sum of their reciprocals ?",
Math. Comp., Volume 61, Number 203, July 1993, pages 117-130.

Rewrite this equation as,

u/v+v+1/u=n

Then, this is birationally equivalent to

y^2+nxy=x^3

by the following birational transformation,

x = -u
y = uv

and

u = -x
v = -y/x