Re: [Cfrg] Question about A=6 Montgomery over 2^89-1
Dan Brown <dbrown@certicom.com> Sat, 12 December 2015 15:23 UTC
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From: Dan Brown <dbrown@certicom.com>
To: Grigory Marshalko <marshalko_gb@tc26.ru>, "cfrg@ietf.org" <cfrg@ietf.org>
Thread-Topic: [Cfrg] Question about A=6 Montgomery over 2^89-1
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Date: Sat, 12 Dec 2015 15:23:25 +0000
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Subject: Re: [Cfrg] Question about A=6 Montgomery over 2^89-1
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So 66^3 is one of the few (13???) integral j-invariants with CM over the rationals (the Baker-Stark-Heegner theorem?). And yes Elkies shows that half the primes will give a supersingular. I'll look at how j=66^3 corresponds to A=6 in 2016, and eventually try to sort out whether there's much to say about small |A|. Original Message From: Grigory Marshalko Sent: Friday, December 11, 2015 4:09 PM To: Dan Brown; cfrg@ietf.org Subject: Re: [MASSMAIL][Cfrg] Question about A=6 Montgomery over 2^89-1 This seems to be a better answer http://alexricemath.com/wp-content/uploads/2013/07/EC2.pdf Regards, Grigory Marshalko, expert, Technical committee for standardisation "Cryptography and security mechanisms" (ТC 26) www.tc26.ru 11 декабря 2015 г., 23:20, "Grigory Marshalko" <marshalko_gb@tc26.ru> написал: > Hi, > > May be this is the case: > from wiki: > If an elliptic curve over the rationals has complex multiplication then the set of primes for which > it is supersingular has density 1/2. If it does not have complex multiplication then Serre showed > that the set of primes for which it is supersingular has density zero. Elkies (1987) showed that > any elliptic curve defined over the rationals is supersingular for an infinite number of primes. > > and this is also may be useful http://pages.cs.wisc.edu/~cdx/ComplexMult.pdf > > Regards, > Grigory Marshalko, > expert, > Technical committee for standardisation "Cryptography and security mechanisms" (ТC 26) > www.tc26.ru > 11 декабря 2015 г., 00:22, "Dan Brown" <dbrown@certicom.com> написал: > >> Hi, >> >> I stumbled upon something surprising (to me), using Sage (while searching >> for something else). >> >> The Montgomery curve y^2 = x^3 + 6x^2 + x over the field of size 2^89-1, has >> order 2^89, so it is maximally vulnerable to Pohlig-Hellman. (Other >> details: it has order p+1, so is also vulnerable to MOV. I haven't checked >> yet, but I'd _bet_ it's supersingular. It has j-invariant 66^3.) >> >> As is well-known, the supersingular curve y^2 = x^3 + x also has order 2^89 >> (it has j-invariant 1728=12^3). But I recall a result of Koblitz saying >> that curves over F_p with order p+1 are very rare (among isomorphism >> classes). Naively, I would think that finding two such curves so close >> together (A=0 and A= 6) has negligible chance, unless these weak curves are >> distributed towards small |A|. >> >> Nonetheless, I still hope that this does _not_ indicate some general _weak_ >> correlation between Montgomery curves with a small coefficient and known >> attacks. >> >> To that end, I'd be curious if somebody here could explain the theory behind >> this example curve. For example, it would be re-assuring to explain this as >> a mere one-time coincidence, rather than a higher chance of a known attack >> (e.g. MOV or PH) on smaller-coefficient curves. (Purely speculating: maybe >> there's a good theory of supersingular j-invariants for each prime p, then a >> way to deduce A from j, such that p=2^89-1 and j=66^3 formed a superstorm to >> arrive at a small A=6.) >> >> Absent such an explanation, the worry is that if known attacks more >> generally exhibit this kind of correlation with coefficient size, then how >> wise is it to suggest small-coefficient curve as a remedy against secret >> attacks? >> >> I am aware that there are other worries of a different nature >> ("manipulation") involved with methods that generate larger coefficients, >> but maybe there's a good way to balance both concerns. >> >> Best regards, >> >> Daniel Brown >> >> _______________________________________________ >> Cfrg mailing list >> Cfrg@irtf.org >> https://www.irtf.org/mailman/listinfo/cfrg
- [Cfrg] Question about A=6 Montgomery over 2^89-1 Dan Brown
- Re: [Cfrg] [MASSMAIL] Question about A=6 Montgome… Grigory Marshalko
- Re: [Cfrg] [MASSMAIL] Question about A=6 Montgome… Grigory Marshalko
- Re: [Cfrg] Question about A=6 Montgomery over 2^8… Dan Brown
- Re: [Cfrg] Question about A=6 Montgomery over 2^8… Grigory Marshalko
- Re: [Cfrg] Question about A=6 Montgomery over 2^8… Paterson, Kenny
- Re: [Cfrg] Question about A=6 Montgomery over 2^8… Ben Laurie
- Re: [Cfrg] Question about A=6 Montgomery over 2^8… Dan Brown