[seqfan] Re: Crypto EllipticK & A000984
Brad Klee
bradklee at gmail.com
Mon Jul 24 21:59:31 CEST 2017
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
More information about the SeqFan
mailing list