# Re: Better key negotiations

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Andrew Del Vecchio wrote:
> That's cool are you working on sample code at this time?
>
> ~Andrew
>
> 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 I have a good idea for key negotiations (NOTE:UNPUBLISHED).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.
>
No. A full proof. Sample code does not matter if it's an insecure
protocol. My presentation needs to be made a lot tighter first.
- --
They who would give up an essential liberty for temporary security,
deserve neither liberty or security
- --Benjamin Franklin
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