Re: [Cfrg] Question about A=6 Montgomery over 2^89-1
Dan Brown <dbrown@certicom.com> Mon, 14 December 2015 03:34 UTC
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From: Dan Brown <dbrown@certicom.com>
To: Steven Galbraith <s.galbraith@math.auckland.ac.nz>, Ben Laurie <ben@links.org>, "Paterson, Kenny" <Kenny.Paterson@rhul.ac.uk>, 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: Mon, 14 Dec 2015 03:34:20 +0000
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Subject: Re: [Cfrg] Question about A=6 Montgomery over 2^89-1
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there's already been some answers on the list, but they haven’t resolved the issue fully to me, so I worked out that j = 256(a^2-3)^3/(a^2-4). For both j and a to be a (rational) integer, we must have a-2 and a+2 divide powers of two, which means a is 0,2,-2,6,-6. I expect this is all worked out in some old publication. so, for other j, I'd expect a mod p not too small. Not sure what happens for irrational j with CM over the rationals. hence, so far, I only expect this one small a example. not yet sure how to infer something from this about unpublished weaknesses in ECDLP (assuming they exist, which is unlikely) Original Message From: Steven Galbraith Sent: Sunday, December 13, 2015 6:26 PM To: Ben Laurie; Paterson, Kenny; Dan Brown; Grigory Marshalko; cfrg@ietf.org Subject: Re: [Cfrg] Question about A=6 Montgomery over 2^89-1 Read Chapters 2 and 4 of Washington's book on Elliptic Curves. Section 4.6 comes close to discussing this sort of thing. For a deeper knowledge you can look at Washington's chapter on Complex Multiplication, or the book by Cox "Primes of the form x^2 + n*y^2", or this great course by Drew Sutherland: http://math.mit.edu/classes/18.783/lectures.html Steven On 14/12/15 11:13, Ben Laurie wrote: > OK, now I want the shortest possible reading course that lets me figure > out what you just said. > > Actually a serious request. > > On Sat, 12 Dec 2015 at 18:29 Paterson, Kenny <Kenny.Paterson@rhul.ac.uk > <mailto:Kenny.Paterson@rhul.ac.uk>> wrote: > > Hi Dan, > > I asked Steven Galbraith about this. Here's what he said: > > "Some thoughts on the elliptic curve: E : y^2 = x^3 + 6x^2 + x > It is a global CM curve with j(E) = 66^3, so it has complex > multiplication of discriminant -16. > Due to congruence conditions it is therefore supersingular. > To flip the question around: For a given prime p, there is usually a CM > elliptic curve over Q which is supersingular modulo p. And since that > curve is from a finite list it usually has small coefficients. So it is > not at all surprising that one can find a curve with small coefficients > that is supersingular. And if the prime p is such that p+1 is smooth > then this is weak for Pohlig-Hellman. > There is no reason to think there would be another other such example, > so I guess there is nothing to worry about." > > Steven no longer subscribes to this list, so include him in cc if > you want > him to see any responses. > > Cheers > > Kenny > > > > On 12/12/2015 15:23, "Cfrg on behalf of Dan Brown" > <cfrg-bounces@irtf.org <mailto:cfrg-bounces@irtf.org> > on behalf of dbrown@certicom.com <mailto:dbrown@certicom.com>> wrote: > > >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 <mailto: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 <http://www.tc26.ru> > >11 декабря 2015 г., 23:20, "Grigory Marshalko" > <marshalko_gb@tc26.ru <mailto: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 <http://www.tc26.ru> > >> 11 декабря 2015 г., 00:22, "Dan Brown" <dbrown@certicom.com > <mailto: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 <mailto:Cfrg@irtf.org> > >>> https://www.irtf.org/mailman/listinfo/cfrg > > > >_______________________________________________ > >Cfrg mailing list > >Cfrg@irtf.org <mailto:Cfrg@irtf.org> > >https://www.irtf.org/mailman/listinfo/cfrg > > _______________________________________________ > Cfrg mailing list > Cfrg@irtf.org <mailto: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