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Re: Better key negotiations

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That's cool are you working on sample code at this time?


Watson Ladd wrote:
> Andrew Del Vecchio wrote:
>> What are "eliptic curves", Watson? I'm not a math master, I just
>> know how to do IT :D
>> ~Andrew
> Elliptic curves are equations of the form y^2=x^3+ax+b. In
> cryptography we consider them over the projective plane formed by a
>  finite field, and we can add points on the curve to form cyclic
> subgroups for which the Diffie-Hellman problem is hard. The main
> advantage is a major speedup, and key sizes can be smaller for the
> same security factor. There are a lot of patents involved, meaning
> you need to pay care to how you are doing the math. But the prize
> is very good security, as no breakthroughs have been made since
> 1985. Check the wiki for details.
>> Watson Ladd wrote:
>>> Jason Holt wrote:
>>>> On Fri, 1 Sep 2006, Watson Ladd wrote:
>>>>> I have a good idea for key negotiations (NOTE:UNPUBLISHED).
>>>>>  Here
>>> it is:
>>>>> Let the server have a public key y=h^x mod p, p=2q+1,
>>>>> h=g^2, and
>>> private
>>>>> key x^-1 mod q, or z. (g is a generator).
>>>>> A client will send y^a and remember a. A server will send
>>>>> back h^b and remember b. The client will compute (h^b)^a.
>>>>> The server will compute (y^a)^(bz). We note that:
>>>>> (y^a)^(bz)=h^(ax*bz)=h^(abxz)=h^(ab)=(h^b)^a, as z and x
>>>>> are multiplicative inverses mod q. We further note that
>>>>> this is just Diffie-Hellman if we replace y with h^z,  a
>>>>> with a*x, and z with 1, b with b. So this is secure if
>>> DDH holds.
>>>>> I am not a cryptographer, so will someone please check this
>>>>>  method. I have not found it anywhere.
>>>> Why would we use this instead of plain-vanilla
>>>> Diffie-Hellman? -J
>>> To authenticate the server to the client. I want to dispense
>>> with RSA as we are putting a critical egg into two baskets at
>>> once. Also, we can migrate to exotic DDH assumption groups if a
>>>  breakthrough happens. Like GF(p^n), n>1, or elliptic curves.

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