[seqfan] Re: Crypto EllipticK & A000984
Andrew N W Hone
A.N.W.Hone at kent.ac.uk
Tue Jul 25 11:35:30 CEST 2017
Dear all,
It has been a long time since I read Edwards' article, and I haven't been following the discussion carefully, but I am familiar with the Edwards curve, as it is one of
a much larger family (so called biquadratic curves) which appear in many different contexts. Apparently Euler may have been the first to study these curves, in his work on
elliptic integrals.
In the family of curves
a = x^2 + y^2 + x^2*y^2
with parameter a, if we view a as a Hamiltonian then Hamilton's equations are the canonical ones
dx/dt = \partial a / \partial y
dy/dt = - \partial a / \partial x
so we find
dx/dt = 2*y*(1+x^2)
dy/dt = -2*x*(1+y^2).
One can show that, for all but finitely many values of a, this extends to a smooth flow on the whole curve a=constant (including points at infinity). This is described, for the case of a general biquadratic curve, in Duistermaat's book "Discrete Integrable Systems: QRT Maps and Elliptic Surfaces".
If I remember correctly, the parametrizations introduced by Edwards can be written using Jacobi elliptic functions.
All the best
Andy
________________________________________
From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of Brad Klee [bradklee at gmail.com]
Sent: 24 July 2017 20:59
To: Sequence Fanatics Discussion list
Subject: [seqfan] Re: Crypto EllipticK & A000984
Hi Sven and All,
Funny insight. As for /decryption/, sometimes it seems like an immediate
research task. A difficult research article is not a ciphertext, but there
is an analogy. It's also important to think for yourself and produce
interesting results. Maybe you would call it /encryption/ if the results
are difficult for a reader to understand.
For example, to someone young and relatively less-educated, it's difficult
to understand even a small part of the original article by Harold Edwards (
cf.
http://www.ams.org/journals/bull/2007-44-03/S0273-0979-07-01153-6/home.html
). Edwards is also constructing parameterized solutions; though, in more
generality.
The approach via polar coordinates and high-school calculus leads to not
just one, but two integrals for the same parameterization. The first is
described above, the second is easier:
* Rewrite the curve in Hamiltonian form: a = x^2 + y^2 + x^2*y^2
* And in polar coordinates: a = r^2 - Z^2*r^4, where Z = 1/2*sin(2*phi) =
cos(phi)*sin(phi)
* Assume real "a" and take the circular solution around surface minimum
a,x,y=0:
r = sqrt( 1/(2*Z^2)*(1 - sqrt( 1 - 4*a *Z^2 ))
* Calculate: dt =dphi* d(r^2)/da = dphi / sqrt(1-4*a*Z^2), as above !
In Hamiltonian mechanics, it's not too difficult to understand why the
a-derivative of the area integral leads to time dependence. In this case,
the addition rules are not the usual Hamilton equations of motion, so the
result seems more surprising.
On this topic, it's also worth mentioning sequences A066770, A066771, which
capture iteration of circular angle addition with each dphi the angle of a
3-4-5 right triangle. Similar integer sequences for the Edwards curve could
be divised.
Thanks,
Brad
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