A Question on Ryley's Theorem
TPiezas
Hello all,
Ryley's Theorem (1825) is that, "Any number N is the sum of three
rational cubes."
Proof: (p^3+qr)^3 + (-p^3+pr)^3 + (-qr)^3 = N (6Nvp^2)^3
where {p,q,r} = {N^2+3v^3, N^2-3v^3, 36N^2v^3}, with v arbitrary.
It has been found that the theorem can be strengthed to "...any
positive number N is the sum of three _positive_ rational cubes", and
can be shown true if v is within a range. For small integral N, the
ranges are:
N = 1: v = {1.32 - 1.42}
N = 2: v = {1.20 - 1.40}
N = 3: v = {2.8 - 3.0}
N = 4: v = {3.4 - 3.7}
and so on. Question: Does anyone know how to calculate v and express
it in terms of N such that all three cubes are positive?
P.S. There _seem_ to be two ranges per N. For example, for N = 2^5,
all terms are positive for a) v = 7-9 and, b) v = 14-15.
- Titus
Kevin Buzzard
TPiezas <tpi...@gmail.com> wrote:
>
> Hello all,
>
> Ryley's Theorem (1825) is that, "Any number N is the sum of three
> rational cubes."
>
> Proof: (p^3+qr)^3 + (-p^3+pr)^3 + (-qr)^3 = N (6Nvp^2)^3
>
> where {p,q,r} = {N^2+3v^3, N^2-3v^3, 36N^2v^3}, with v arbitrary.
>
> It has been found that the theorem can be strengthed to "...any
> positive number N is the sum of three _positive_ rational cubes", and
> can be shown true if v is within a range. For small integral N, the
> ranges are:
>
> N = 1: v = {1.32 - 1.42}
> N = 2: v = {1.20 - 1.40}
> N = 3: v = {2.8 - 3.0}
> N = 4: v = {3.4 - 3.7}
>
> and so on. Question: Does anyone know how to calculate v and express
> it in terms of N such that all three cubes are positive?
I didn't think about the specific question you ask, but let me just
make the following fun remark. If a rational number t>0 is of the form a^3-b^3
with a,b>=0, then it's of the form c^3+d^3 with c,d>=0.
This is because if you think about the curve X^3+Y^3=t as an elliptic
curve with point at infinity at the origin, then the real points are
isomorphic (as group) to a circle, and the region with X,Y>=0 corresponds
to e^{i.theta} with 2*pi/3<=theta<=4*pi/3 (proof: points of inflexion).
In practice you can find c and d by repeated doubling (i.e. drawing the
tangent at (a,-b) and continuing with this until you succeed; the argument
above shows you will).
Hence if you have a solution for N=x^3+y^3+z^3 with N>0, at least
one of x,y,z will be >=0, and applying this algorithm (drawing tangents) will
eventually make two and then three of them >=0.
Laurent Bartholdi
On Nov 10, 6:49�pm, TPiezas <tpie...@gmail.com> wrote:
> Hello all,
>
> Ryley's Theorem (1825) is that, "Any number N is the sum of three
> rational cubes."
>
> Proof: (p^3+qr)^3 + (-p^3+pr)^3 + (-qr)^3 = N (6Nvp^2)^3
>
> where {p,q,r} = {N^2+3v^3, �N^2-3v^3, 36N^2v^3}, with v arbitrary.
>
> It has been found that the theorem can be strengthed to "...any
> positive number N is the sum of three _positive_ rational cubes", and
> can be shown true if v is within a range. �For small integral N, the
> ranges are:
>
> N = 1: �v = {1.32 - 1.42}
> N = 2: �v = {1.20 - 1.40}
> N = 3: �v = {2.8 - 3.0}
> N = 4: �v = {3.4 - 3.7}
>
> and so on. �Question: Does anyone know how to calculate v and express
> it in terms of N such that all three cubes are positive?
It seems easy to express v in terms of N, just by solving the above
equations for v. Setting (to ease notation) x=N^2 and y=v^3, one gets
the constraints
x-3y < 0
x^2 - 30xy + 9y^2 < 0
x^3 + 45x^2y - 81 y^2 + 27y^3 > 0
which are linear (!) in x,y -- they define 6 lines in the {x,y} plane
delimiting, as you noted, two regions. Numerically, these regions are
roughly
y/x in [1/3,0.764] or [2.253,3.300].
TPiezas
On Nov 11, 5:48�am, Laurent Bartholdi <laurent.bartho...@gmail.com>
wrote:
>
> - Show quoted text -
Hello Laurent,
Never mind my question about whether 2.253... and the others are
algebraic numbers. I just realized that 0.7684... (there was a typo)
and 2.253... are the positive roots (rounded off) of 1+45u-81u^2+27u^3
= 0 and
3.299... is the larger root of 1-30t-9t^2 = 0.
I've updated my site on Algebraic Identities with your analysis.
Kindly see Form 2: x^3+y^3+z^3 = N here, http://sites.google.com/site/tpiezas/001
- Titus
me13013
On Nov 10, 6:49?pm, Kevin Buzzard <buzz...@imperial.delete.ac.uk>
wrote:
> I didn't think about the specific question you ask, but let me just
> make the following fun remark. If a rational number t>0 is of the form a^3-b^3
> with a,b>=0, then it's of the form c^3+d^3 with c,d>=0.
> This is because if you think about the curve X^3+Y^3=t as an elliptic
> curve with point at infinity at the origin, then the real points are
> isomorphic (as group) to a circle, and the region with X,Y>=0 corresponds
> to e^{i.theta} with 2*pi/3<=theta<=4*pi/3 (proof: points of inflexion).
> In practice you can find c and d by repeated doubling (i.e. drawing the
> tangent at (a,-b) and continuing with this until you succeed; the argument
> above shows you will).
>
> Hence if you have a solution for N=x^3+y^3+z^3 with N>0, at least
> one of x,y,z will be >=0, and applying this algorithm (drawing tangents) will
> eventually make two and then three of them >=0.
Kevin,
a^3-b^3 = c^3+d^3 has been discussed recently in alt.math.recreational
(the thread is http://tinyurl.com/ybouceg ). There I present an
outline of a proof that I think is equivalent to repeated doubling,
but there may be a couple holes in my proof (I'm more of a hobbyist
than a mathematician). Could you look at what I posted there and tell
me if it holds water?
And the original poster there is looking for some history of this
problem. From your post I gather you've seen this problem before. Do
you anything about the history of this problem?
Bob H
Kevin Buzzard
me13013 <me1...@gmail.com> wrote:
>
> a^3-b^3 = c^3+d^3 has been discussed recently in alt.math.recreational
> (the thread is http://tinyurl.com/ybouceg ). There I present an
> outline of a proof that I think is equivalent to repeated doubling,
> but there may be a couple holes in my proof (I'm more of a hobbyist
> than a mathematician). Could you look at what I posted there and tell
> me if it holds water?
You've had the key idea: drawing the tangent. The crucial point is
that if you keep doing this, then you will succeed. Why? I said "general
theory of elliptic curves shows you will" but let me be a bit more
precise. General theory of elliptic curves tells us that there is
a continuous map from the real points on curve x^3+y^3=c (c>0) to the
nearly-a-circle e^{i.theta}, 0<theta<2.pi, with the following properties:
1) if x-->infty and y-->-infty on the curve, then theta-->0 from below
on the circle
2) if x-->-infty and y-->infty on the curve, then theta-->0 from above
on the circle
and so let's "complete" the picture by sending "infinity" on the curve
to theta=0 on the circle. Let me call this map "P goes to theta(P)". Now
3) [crucial] given a point (x,y) on the curve, if you draw the tangent
and compute the 3rd point of intersection as (x2,y2), then on the curve
this corresponds to the map sending theta to -2*theta (mod 2.pi),
i.e. theta(x2,y2)=-2*theta(x,y).
This is because secretly the map I'm talking about is a group homomorphism,
and is in some sense where the work is. In fact more generally if P and Q
are two points on the curve, and R is the 3rd point of intersection of
the line PQ and the curve, then theta(P)+theta(Q)+theta(R)=0 mod 2pi.
And now to finish, (3) implies
4) the points x=0 and y=0 on the curve correspond to the points
theta=2*pi/3 and theta=4*pi/3
(because x=0 and y=0 on the curve are the two points where the
chord-tangent method gets you back to where you started, and
those two values of theta are the ones fixed by theta-->-2*theta.
So now you just need to check that for any theta not equal to 0 mod 2*pi,
if you keep multiplying by -2 then eventually you'll land in the region
between 2*pi/3 and 4*pi/3. This is not hard.
And that's the way I was thinking to make the argument rigorous.
> And the original poster there is looking for some history of this
> problem. From your post I gather you've seen this problem before. Do
> you anything about the history of this problem?
Hendrik Lenstra told me the problem in the 1990s, and I found the solution
above, and Lenstra told me that it was also the solution he found.
I am no historian but I believe that there's explicit evidence that
Diophantus knew that if he had 2 points on a plane cubic he could
construct a 3rd in this way, so it wouldn't surprise me if his
proof was close to this one.
Kevin