Diophantus' Lemma
James Dow Allen
It must be quite old ... in fact 17 centuries old!
For any given positive rational numbers a,b
with a > b
Find positive rational numbers c,d
such that
a^3 - b^3 = c^3 + d^3
The result is one of the Lost Lemmas of
Diophantus of Alexandria: he cites it in his
_Arithmetica_ but the section that contains
his proof is lost.
http://en.wikipedia.org/wiki/Diophantus#The_Porisms
There's a nice website with such results:
http://sites.google.com/site/tpiezas/010
Diophantus' Lemma may be at that site somewhere,
but of course the challenge is to rediscover from scratch.
This Lemma may show why Diophantus is so highly
respected. Does he have any other outstanding
lemmas?
James Dow Allen
me13013
> For any given positive rational numbers a,b
> with a > b
> Find positive rational numbers c,d
> such that
> a^3 - b^3 = c^3 + d^3
I think the approach might go something like this (this parrots the
usual method for addition on an elliptic curve):
First, divide by b, so the problem is reduced to finding x,y s.t. x^3
+ y^3 = z^3 - 1, where z=a/b is rational > 1 and x,y will be rational
> 0.
It is useful to think of this as a curve in the x,y plane, with
x^3+y^3=k (k=z^3-1 is a constant). Ignoring the constraint that x,y
be positive, we have one rational point on this curve, namely x=z and
y=-1. Consider the tangent line to the curve at this point. This
line has rational slope (technically I would have to consider the case
that it has infinite slope, or show that it doesn't) because the
derivative at this point is rational.
The equation describing the intersection of this line and the curve
will be a cubic, so it has exactly three roots. Two of these roots
will correspond to the point. The third root will give us a point on
the curve, and hopefully it will not be the same point (which would be
a triple root). That point will necessarily be rational, since the
cubic can't have two rational roots and one irrational. What I don't
know is whether that point will be positive.
Bob H
me13013
Here's an example to show the flavor of how this would work. Suppose
we start with a=13, b=10. So our target is 13^3 - 10^3 = 1197.
Reducing by dividing by 10^3, we are looking for x,y s.t. x^3 + y^3 =
1197/1000.
We know one point on this curve, x=13/10, y=-1. The slope of our
curve at any point is -x^2/y^2. So the tangent line at our known
point has slope -169/100 and the equation for the line is y+1/
(x-13/10) = -169/100. This simplifies to y = (1197-1690x)/1000.
The intersection of the line and the curve will thus have x that
satisfy x^3 + ((1197-1690x)/1000)^3 = 1197/1000. Collecting terms
(and dividing by the coffeicient of x^3) reveals a cubic of the form
x^3+ux^1+vx-w = 0. We know the three roots will sum to w, and since
we know two of the roots are 13/10, we can easily compute the third
root from w. I cheated and used mathematica. The third root is
2561/31970. The corresponding point has x=2561/31970, y=3394/3197.
So (2561/31970)^3 + (3394/31970)^3 = (13/10)^3 - 1.
Multiplying all terms by 10^3, we get
(2561/3197)^3 + (33940/3197)^3 = 13^3 - 10^3.
In this case the third point on the curve has both points positive, so
it gives us a solution. I'm not sure if that will always be the
case. If not, we could then apply the same process at the new point
to get a third point. So even if one application of the process can't
be shown to always produce a positive point, possibly some hill-
climbing argument could show that repeated application will always
lead to one.
Bob H
me13013
Working that through symbolically ... I get ... x=(z^4-2z)/(z^3+1) and
y=(2z^3-1)/(z^3+1) are such that x^3 + y^3 = z^3 - 1.
However, this will not always lead to a solution in positive numbers.
We have z>1, and it can be shown that y is strictly positive for such
z. However, x will be positive only when z>2^(1/3), which is about
1.26. Amazing that I picked an example so close to this cutoff.
So, not quite to a valid solution, is it?
Bob H
James Dow Allen
> > > On Nov 7, 3:12 am, James Dow Allen <jdallen2...@yahoo.com> wrote:
> > > > For any given positive rational numbers a,b
> > > > with a > b
> > > > Find positive rational numbers c,d
> > > > such that
> > > > a^3 - b^3 = c^3 + d^3
>
> y=(2z^3-1)/(z^3+1) are such that x^3 + y^3 = z^3 - 1.
Thank you, Bob!
I'm glad :-) you confirmed that this is a relatively
difficult problem; I was feeling blue to have made
no progress. :-(
What seems *really* remarkable is that the ancient
Diophantus apparently found such a result.
His proof is lost, but it would be nice to see his
mention first-hand. (All I've found on this matter
are brief second-hand mentions at sites like
Wikipedia.)
I like integers, so converted your formula to
an integer-only form (z = a/b):
(a^4 - 2*a*b^3)^3 + (2*b*a^3 - b^4)^3
= (a^3 - b^3) * (a^3 + b^3)^3
Of course it suffers from the same problem as
yours: a^4-2*a*b^3 < 0 when a/b < 1.2599.
There are certainly solutions *sometimes* when
z < 1.2599, for examples:
6^3 - 5^3 = 4^3 + 3^3
9^3 - 8^3 = 6^3 + 1^3
10^3 - 9^3 = (8649^3 + 1999^3) / 1342^3
12^3 - 11^3 = (360^3 + 37^3) / 49^3
On the other hand, brute force searches when
z = 5/4 or z = 8/7 suggest that those solutions
must have very large denominators.
Thanks again, Bob. As I say, my real interest is
in Diophantus. Did he really solve this?
Was he some kind of Wunderkind?
James
me13013
It may interest you to know that the approach I took is similar to one
I learned for solving quadratics in rationals, and that approach I
learned from "Diophantus and Diophantine Equations" (by Bashmakova).
That book does have some info on solving certain cubics but I did not
understand it at the time I read it (about 10 years ago). More
recently I've been reading about elliptic curve cryptography, which
deals with finding rational points on curves of the form y^2 = x^3 +
ax^2 +bx + c. The common thread with all this is that if you have one
rational point on a curve you can use it to find others. Also, if you
have two rational points on a cubic you can find a third, but I did
not make use of that fact. For all I know, that approach may have
been invented by Diophantus.
Expanding my formula back to the original problem statement a^3 - b^3
= c^3 + d^3, I get c=(2ba^3-b^4)/(a^3+b^3) and d=(a^4 - 2*a*b^3)/
(a^3+b^3). Here I've swapped c and d because if one of them is
negative, I want it to be d.
Now, if d is negative, we can just apply my formula again. For
example if a=6 and b=5 we have (1535/341)^3 + (-204/341)^3 = a^3 -
b^3. The left side can be converted to (1535/341)^3 - (204/341)^3.
Treating these as a' and b', we can apply the formula again to reveal
that (1473924701544/1236225608299)^3 + (5525732982145/1236225608299)^3
= a'^3 - b'^3 = a^3 - b^3 = 6^3 - 5^3 = 91.
That was what I was alluding to when I said something about hill-
climbing. Applying the formula a second time might not produce an
answer in positive values either (I haven't checked, maybe it always
does). But it might be possible to show that continued application of
the formula will eventually produce a suitable answer. We also don't
know if Diophantus' proof was constructive. He may have just proven
that a solution always exists, without providing any method to find
it.
This do-it-twice method reveals
5^3 - 4^4 = (152529295/960946371)^3 + (3782680016/960946371)^3
and
8^3 - 7^3 = (3676875939616/1171919091705)^3 +
(6057752297009/1171919091705)^3
I'll agree that those denominators are pretty large for brute search.
There may be smaller denominators that work though.
Bob H
me13013
The do-it-twice method should work for 1.029 < z < cube root(2)
(1.029 is approximate). Between 1 and 1.029 we still get a negative
value. For example, do-it-twice for 106/103 fails. But do-it-THRICE
does succeed (with a 131-digit denominator that I'm not going to
bother posting). But then do-it-thrice fails in the interval 1 to
1.0035.
I guess one could keep going, but at that point you're dealing with a
formula for z that is order 64 in the numerator. There's bound to be
some way to show that with enough of these steps you'll eventually get
to a positive solution.
Bob H
James Dow Allen
> The do-it-twice method should work for 1.029 < z < cube root(2)
> (1.029 is approximate). Between 1 and 1.029 we still get a negative
> value. For example, do-it-twice for 106/103 fails. But do-it-THRICE
> does succeed (with a 131-digit denominator that I'm not going to
> bother posting). But then do-it-thrice fails in the interval 1 to
> 1.0035.
I've exchanged e-mail with Tito Piezas who's already editing his
page
http://sites.google.com/site/tpiezas/012
to incorporate some of your work.
> > > What seems *really* remarkable is that the ancient
> > > Diophantus apparently found such a result.
> > > His proof is lost, but it would be nice to see his
> > > mention first-hand. (All I've found on this matter
> > > are brief second-hand mentions at sites like
> > > Wikipedia.)
Is there a history-of-math group? Did Diophantus
really figure this out, or was it just a wild guess?
James
me13013
> I've exchanged e-mail with Tito Piezas who's already editing his
> page
> http://sites.google.com/site/tpiezas/012
> to incorporate some of your work.
I think with a little more thought we can make a valid proof.
First, the z=13/10 case is shown in this graph:
http://bumblebeagle.org/miscellany/13_over_10.png
Using the tangent line at some rational point (x,y) produces a new
solution point ( x(x^3+2y^3)/(x^3-y^3) , y(y^3+2x^3)/(y^3-x^3)). If x
> -y > 0, then (I claim) the second coordinate in the new point will
always be positive. If the first coordinate is also positive we are
finished. If it is negative we swap the coordinates and try again.
See
http://bumblebeagle.org/miscellany/11_over_10.png
for an example of trying again.
So basically we have a transformation of (x,y) to ( y(y^3+2x^3)/(y^3-
x^3) , x(x^3+2y^3)/(x^3-y^3)). (Note that I've swapped coordinates).
All that is needed is to show that repeated application of this
transformation, given valid starting conditions (x > -y > 0), will
eventually produce a point with both coordinates positive. The first
point is (x,y) = (z,-1).
The two plots suggest this is true. If it can be shown that (a) the
transformation (x,y) to (x',y') either produces both coordinates
positive, or produces x' > -y' > 0 (i.e. it maintains the valid
starting condition) and (b) y' > y, then either y* will eventually
become positive, or y* will approach some limiting value less than or
equal to zero. By y* I mean the sequence of y values as we repeatedly
apply this transformation.
I think it is easy to show (a) and (b) because of the slope of the
curve. I'm not sure how to show that y* doesn't have some limiting
negative value.
Hopefully I am making sense,
Bob H
TPiezas
Hello all,
Hm, interesting mathgroup. Anyway, the original problem by J. Allen
was,
1. "For any given positive rational numbers {a,b} (with a > b), find
positive rational numbers {c,d} such that a^3 - b^3 = c^3 + d^3."
If we modify it slightly to, "...find positive rational numbers
{c,d,e} such that a^3 - b^3 = c^3 + d^3+ e^3", then there is even a
_formula_ to do it.
Pls see Form 2, Ryley's Theorem, at http://sites.google.com/site/tpiezas/001.
I recently updated it yesterday by the formula given by someone from
sci.math.research who was kind enough to analyze the problem.
- Titus
me13013
> Anyway, the original problem by J. Allen was,
>
> 1. "For any given positive rational numbers {a,b} (with a > b), find
> positive rational numbers {c,d} such that a^3 - b^3 = c^3 + d^3."
>
> If we modify it slightly to, "...find positive rational numbers
> {c,d,e} such that a^3 - b^3 = c^3 + d^3+ e^3", then there is even a
> _formula_ to do it.
>
> Pls see Form 2, Ryley's Theorem, at http://sites.google.com/site/tpiezas/001.
It's not immediately clear to me how we can apply Ryley's Thm to solve
the original problem. The original problem can be restated as given
{a,b} find {c,d} s.t. a^3 = b^3 + c^3 + d^3 (all in positive rationals
with a>b). Ryley's Thm can partition a^3 into three cubes (in
infinitely many ways, it seems), but there is no guarantee that one of
them is b^3. And I can't see how I could work backwards to partition
some other value into three cubes, scale so that one of those three is
b^3, and then end up with a^3 as the sum.
I haven't looked at the thm in detail though. Perhaps I can always
pick v s.t. one of the three cubes is b^3. Is that what you are
proposing?
Bob H
me13013
> sci.math.research who was kind enough to analyze the problem.
James, you might want to look at that thread at sci.math.research
( http://tinyurl.com/yglzgt7 ). Kevin Buzzard mentions the a^3-b^3 =
c^3+d^3 problem and outlines a solution method that is probably
equivalent to the one I came up with. Only, he sounds like he knows
what he is talking about.
Bob H
Chip Eastham
> On Nov 12, 8:54 am, TPiezas <tpie...@gmail.com> wrote:
>
> > I recently updated it yesterday by the formula given by someone from
> > sci.math.research who was kind enough to analyze the problem.
>
> James, you might want to look at that thread at sci.math.research
> c^3+d^3 problem and outlines a solution method that is probably
> equivalent to the one I came up with. Only, he sounds like he knows
> what he is talking about.
>
> Bob H
To summarize a bit, Bob gives a formula which
is similar (equivalent) to three given by
Vieta (_Zetetica_ IV, problems 18-20, 1591):
Given rational a > b > 0, find rational x,y
such that:
1. a^3 - b^3 = x^3 + y^3; satisfied by
x = a(a^3 - 2b^3)/(a^3 + b^3)
y = b(2a^3 - b^3)/(a^3 + b^3)
2. a^3 + b^3 = x^3 - y^3; satisfied by
x = a(a^3 + 2b^3)/(a^3 - b^3)
y = b(2a^3 + b^3)/(a^3 - b^3)
3. a^3 - b^3 = x^3 - y^3; satisfied by
x = b(2a^3 - b^3)/(a^3 + b^3)
y = a(2b^3 - a^3)/(a^3 + b^3)
This account may be found in Heath's now
century old English translation of the
six books of Arithmetica then known,
_Diophantus of Alexandria_, Chapter V (an
introduction to Book V), pp. 101-102, a
PDF of which may be downloaded here (~18Mb):
[Public domain materials in ancient math]
http://www.wilbourhall.org/
Heath says that Fermat observed that
in precisely the case (1) fails to give
positive x,y, namely a/b < cuberoot(2),
the case (3) succeeds in giving another
pair of positive rationals whose cubes
are the same difference as a^3 - b^3,
"so that by employing them successively
we can effect the transformation
required ... even when a^3 is not > 2b^3."
We can analyze this more fully by noting
that the homogeniety of formulas in (3)
allows mapping from the ratio r = a/b to
a new ratio f(r) = x/y:
f(r) = (b(2a^3 - b^3))/(a(2b^3 - a^3))
= (2r^3 - 1)/(r(2 - r^3))
The function f(r) has a fixed point at
r = 1, but is easily seen to be positive
and increasing on (1,cuberoot(2)):
f'(r) =
(6r^2)(r(2-r^3)) - (2 - 4r^3)(2r^3-1)
-------------------------------------
r^2 (2-r^3)^2
= 6r/(2-r^3) + 2[f(r)]^2
Consequently f'(r) > 8 on (1,cuberoot(2),
so that above fixed point f(1) = 1 and
below the singularity at r = cuberoot(2)
the iteration r |-> f(r) will move at
least 8 times the distance above 1 at
each repetition.
While there is no uniform upper bound on
how many repetitions may be required to
get r > cuberoot(2), a bound may be given
in terms of r - 1. Specifically, if k
is the least iteration s.t. f^k(r) will
exceed cuberoot(2), we know:
k <= [log_2(cuberoot(2)-1) - log_2(r-1)]/3
regards, chip
Chip Eastham
>
> > I recently updated it yesterday by the formula given by someone from
> > sci.math.research who was kind enough to analyze the problem.
>
> James, you might want to look at that thread at sci.math.research
> c^3+d^3 problem and outlines a solution method that is probably
> equivalent to the one I came up with. Only, he sounds like he knows
> what he is talking about.
>
> Bob H
To summarize a bit, Bob gives a formula which
Chip Eastham
> While there is no uniform upper bound on
> how many repetitions may be required to
> get r > cuberoot(2), a bound may be given
> in terms of r - 1. Specifically, if k
> is the least iteration s.t. f^k(r) will
> exceed cuberoot(2), we know:
>
> k <= [log_2(cuberoot(2)-1) - log_2(r-1)]/3
My bound requires a ceiling, since k must
be a positive integer # of repetitions:
Should be:
k <= ceil([log_2(cuberoot(2)-1) - log_2(r-1)]/3)
regards, chip