R.x == d.x |if R.x equal to Publickey x we can solve

[same2026]

k*G=r.x  == d*G=d.x

My English littlebit, but I was told my concept.

if R equal to Publickey x  we can solve

this is example only
R                   = 0x7362f4d6ce57b70f0fc9781dbc92010f709ea60bee3b43bdfd763e860d6eff2f
S                   = 0xbb5739e3279db95257555a571a8a8c1cb36be86c1423b92a46137afa0600fceb
Z                   = 0x119c63b888b501a28c1e3902b65a8499a7d1bddcbc14120906c1e00a64f7ada5
Publickey    = 0x7362f4d6ce57b70f0fc9781dbc92010f709ea60bee3b43bdfd763e860d6eff2f, f1f4569c03bc6c0defa5a84b8ebe26a433504a0847b971aa55c8071b7f36ba18


i can prove if R.x==Publickey.x .if any one share one valid R,S,Z  i send Private-key  challenge me

same_must = 0x12345  # change your key 2**256 or any range
nonce = same_must
d          = same_must
z          = random(2**256)

send me R,S,Z , then i was crack then send
#2nd
Does anyone know of this type of puzzle case? If you do, please share the TXID or address here

[ercewubam]

It can be trivially solved:

Code:

s=(z+rd)/k
d=k
s=(z+rk)/k
s=(z/k)+r
s-r=z/k
(s-r)/z=1/k
k=z/(s-r)

Which means, that if you have z-value, r-value, and s-value, then you can calculate just "z/(s-r)", and recover the private key in this way. It is unlikely, that there are any such transactions in the wild, because anyone could easily sweep these coins.