The prize was offered in this thread:
https://bitcointalk.org/index.php?topic=5321454.0Unfortunately, he locked the thread before I could claim the prize
but I have the solution for the prize !!!
From each message we can derive the z value (hash of the message) so:
First message and signature (m, r, s, z)
Second message and signature (m', r', s', z')
Therefore: ks = z + rd
A and k's' = z' + r'd
ATherefore: (sk - z)/r = (s'k' - z')/r'
But in this case k' = 2k so:
(sk - z)/r = (2s'k - z')/r'
So all you have to do is solve for k. All the other values: s, z, r, s', z', and r' are all known.
rr'(sk - z)/r = rr'(2s'k - z')/r'
r'(sk - z) = r(2s'k - z')
r'sk - r'z = 2rs'k - rz'
r'sk - r'z +rz' = 2rs'k
k = (r'sk - r'z +rz')/2rs'
Once you know k you can simply calculate the private key, d
A = (sk - z)/r
Please send my 3,350 bitcoins to
1BurtWEejbnKeBRsvcydJvsNztB1bXV5iQ
That was easy! Thanks!
Easiest $170,000,000.00 I have ever earned.
Still waiting on payment.