Post
Topic
Board Development & Technical Discussion
Re: Pollard's kangaroo ECDLP solver
by
brainless
on 18/07/2023, 06:03:02 UTC
 Who can solve
 p = 115792089237316195423570985008687907852837564279074904382605163141518161494335
a = 1099511627776
b = 115792089237316195423570985008687907852837564279074904382605163141005436653346
c = (a-b) %p
result = 1612236468765

in pubkey

p =115792089237316195423570985008687907852837564279074904382605163141518161494335
a = 02feea6cae46d55b530ac2839f143bd7ec5cf8b266a41d6af52d5e688d9094696d
b = 02746bd76e07a0dbbcc610245439ee1db94f73b70df43bc543d4046ebe119ad6b3
c = (a-b) %p
result = 02b21dd66bfde832c2dae35688c0e15b91b274ec018e2c14e23f1ca7cb32fcca73

substract formula
p = int(2**256 - 2**32 - 977)
x1 =  # fill pubkey1-x
y1=  # fill pubkey1-y
x2=  # fill pubkey2-x
y2=  # fill pubkey2-y



dx = (x1 - x2) % p
dy = (y1 - (-y2)) % p
c = dy * gmpy2.invert(dx, p) % p
Rx = (c*c - x2 - x1) % p
Ry = (c*(x2 - Rx) - y2) % p
print (Rx , Ry)
print (hex(Rx) , hex(Ry))


if you have alternate formula for adjust with mod p, apply and check for get acurate result in pubkey