two-factor authentication problems
Gabriel Haythornthwaite
gabriel at castelain.com.au
Sun Mar 6 23:51:07 EST 2005
You're quite correct Matt,
Which is why IMHO you can only really get true "non-repudiation" through use
of PKI. And more specifically:
- where the key pair was generated by the end-user, and
- where the server has stored a copy of the transaction - digitally signed
by the end user - which it can reproduce in court.
In this case, a corrupt operator could not have faked the transaction even
if they had wanted to.
RSA SecureID and OATH technology have some great virtues:
- they cost nothing to integrate at the client end - there is no client
"footprint" so there's nothing to go wrong
- they are relatively easy to understand and use
- they're unquestionably better than reliance on user IDs and passwords.
What they won't do is:
- provide non-repudiation
- defend against man-in-the-middle attacks, or
- provide a robust defence against phishing scams
And for these reasons I suspect their days are numbered.
All the best,
Gabriel Haythornthwaite
gabriel at castelain.com.au
Phone: +61 412 544 632
Fax: +61 2 9798 3935
www.castelain.com.au
> -----Original Message-----
> From: owner-cryptography at metzdowd.com
> [mailto:owner-cryptography at metzdowd.com] On Behalf Of Matt Crawford
> Sent: Monday, 7 March 2005 1:38 PM
> To: Ed Gerck
> Cc: cryptography at metzdowd.com
> Subject: Re: two-factor authentication problems
>
>
>
> On Mar 5, 2005, at 11:32, Ed Gerck wrote:
>
> > The worse part, however, is that the server side can always
> fake your
> > authentication using a third-party because the server side
> can always
> > calculate ahead and generate "your next number" for that
> third-party
> > to enter -- the same number that you would get from your
> token. So, if
> > someone breaks into your file using "your" number -- who is
> > responsible? The server side can always deny foul play.
>
> Huh? The server can always say "response was good" when it wasn't
> good. Unless someone reclaims the server from the corrupt
> operator and
> analyzes it, the results are the same.
>
>
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