Re: [Cfrg] Preliminary disclosure on twist security ...

Watson Ladd <> Wed, 26 November 2014 18:05 UTC

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Date: Wed, 26 Nov 2014 10:05:32 -0800
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From: Watson Ladd <>
To: Dan Brown <>
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Subject: Re: [Cfrg] Preliminary disclosure on twist security ...
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On Wed, Nov 26, 2014 at 8:49 AM, Dan Brown <>; wrote:
>> -----Original Message-----
>> From: Watson Ladd
>> The patent in question is US6563928.
>> The claim cited reads as follows:
>> "53. A method of establishing a session key for encryption of data between a
>> pair of correspondents comprising the steps of one of said correspondents
>> selecting a finite group G, establishing a subgroup S having an order q of the
>> group G, determining an element α of the subgroup S to generate greater than
>> a predetermined number of the q elements of the subgroup S and utilising said
>> element α to generate a session key at said one correspondent."
>> "59: 58. A method according to claim 53 wherein said order of said subgroup is
>> of the form utilising an integral number of a product of a plurality of large
>> primes.
>> 59. A method according to claim 58 wherein the order of said subgroup is of
>> the form nrr′ where n, r and r′ are each integers and r and r′ are each prime
>> numbers."
>> This doesn't appear to have anything to do that directly with twist security.
> Well, this is what I was thinking:
> Let F_p be the underlying field.
> Let E be the twist-secure curve, with size #E(F_p) = hr, where h is a small cofactor and r a large prime.  Its twist E' has size h'r' where h' to the another small cofactor and r' is another large prime.
> Now G be the group of F_p^2 rational points, which is a group of size hh'rr', right?

Nope: Take t=p+1-hr. t is the trace of a matrix with determinant p,
say diagonal with \alpha and \beta as eigenvalues. |G| = p^2+1-t_2,
where t_2=\apha^2+\beta^2. Using Viete's formulas, or maybe Newton's,
we write t^2-2p=t_2. So the order of |G| is p^2+2p+1-(p+1-hr)^2. It's
not hh'rr'.

I may have made a typo in the above: check Silverman for the exact details.

>Then let S be the subgroup with of G of size q = rr'.
> Let alpha be the element of S used to generate the yet smaller subgroup of size, i.e. the conventional DH prime-order subgroup of E(F_p).  Now alpha generates r elements, which is greater than a predetermined number, e.g. 2^250.
> This means putting n =1 in Claim 59.
> Best regards,
> Dan

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