X3+Y3=A (A ∈Z; X, Y ∈Q) |
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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)/6xand
Y = (36A+y)/6xx = 12A/(X+Y)
y = 36A(Y-X)/(X+Y).More generally, the curve,
au3 + bv3 + cw3 = 0is equivalent to the following curve,r3 + s3 + abct3 = 0by the following transformation,r = - 6bc2v3w6 - c3w9 - 3b2cv6w3 + b3v9and it is equivalent to
s = - 3bc2v3w6 + c3w9 - 6b2cv6w3 - b3v9
t = - 3uvw (b2v6 + bcv3w3 + c2w6)y2 = x3 - 432a2b2c2by the following transformation,x = 4 (b2v6 + bcv3w3 + c2w6) / u2v2w2
y = 4 (2b3v9 + 3b2cv6w3 - 3bc2v3w6 - 2c3w9) / u3v3w3
Y2=DX4+1 |
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This curve is birationally equivalent to,
E : y2 = x3 - 4Dxunder the following birational transformation,x = 2(Y+1) / X2and
y = 4(Y+1) / X3X = 2x / y
Y = - 1 + 2x3/y2.More generally, the curve,
v2 = au4 + bu3 + cu2 + du + q2is birationally equivalent to,y2 + a1x + a3y = x3 + a2x2 + a4x + a6under the following birational transformation,x = (2q(v+q)+du) / u2and
y = (4q2(v+q)+2q(du+cu2)-d2u2/2q) / u3a1 = d/q
a2 = c - d2 / 4q2
a3 = 2qb
a4 = -4q2a
a6 = a2a4 = a(d2-4q2c)
Congruent Number |
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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 byJ. B. Tunell, "A Classical Diophantine Problem and Modular Forms of Weight 3/2", Inventiones mathematicae, 72(1983)323-334.
Concordant Numbers |
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If there exists the solution of following two quadratic forms simultaneously,
x2 + y2 = z2n is called as concordant. If n is concordant, the following elliptic curve,
x2 + ny2 = w2E : 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 ba=v/(n+u)when
b=(n+u)(2un+n+u2)/(n+u)2a=p/qcompute x and y by p and q such asx=p2-q2then x2+y2 and x2+n*y2 are perfect square.
y=2pq
Let z and w byz=sqrt(x2+y2)then x, y, z and w satisfy
w=sqrt(x2+n*y2)x2 + y2 = z2
x2 + ny2 = w2
(x+y+z)(1/x+1/y+1/z)=n (n, x, y, z ∈Z) |
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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)=nxyzwhere
x2(y+z)+x(y2+(3-n)yz+z2)+yz(y+z)=0
x={-y2+(n-3)yz-z2-Δ}/2(y+z)
Δ2=y4-2(n-1)y3z+(n2-6n+3)y-2z-2-2(n-1)yz3+z4This is birationally equivalent to the elliptic curve
τ2=σ(σ2+(n2-6n-3)σ+16n)and
σ=-4(yz+zx+xy)/z2
τ=2(σ-4n)y/z-(n-1)σx,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.
x/y+y/z+z/x=n (n, x, y, z ∈Z) |
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Rewrite this equation as,
u/v+v+1/u=nThen, this is birationally equivalent toy^2+nxy=x^3by the following birational transformation,x = -uand
y = uv
u = -x
v = -y/x
table of curves (English) |
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