Post
Topic
Board Bitcoin Discussion
Re: Bitcoin puzzle (3,350.00 BTC's )
by
interiawp
on 21/07/2021, 17:16:54 UTC
According to question from bytcoin (first page):

@bytcoin:

calculate k2 when k2 is 2xk1 with the same private key is trivial.

but in your example:

Nonce k1
r = 385570073629551546729230374184391439442417095537024350206131291065055332733
s = 76130260662571134678372154009160868461537800596554110575850229341703072615102
h = 42268864623620244998249895290537358742158169262361846431272029622156081203721

Nonce k2 (k1 / 2 = k2)
r = 79948557280043978274449002532600374620672215928697101734828211284733327158664
s = 56368220214898939855776145385640746541620815395185097101215290078962069907292
h = 87726031595873281985439431891132599611815665254308710544703498112157039462043

one of this h (hash of message) is wrong! this is why it cannot  nonce be calculated.

it cannot be calculated, becouse: you have divide by application real (r,s,h) to get 1/2 of k.

your program cannot calculate 1/2 h > "hash of message" properly.
and this is why others calculation(result) are wrong.

if you give me real values( 2 transaction ) with information that k1 = 2 x k2 or k1 = 10x k2 or whatever
I will give you the nonce k1 and k2 and key.

If you want just pm.