Quote from: nc50lc on Today at 08:13:28 AM

Quote from: weyrfencing18 on November 16, 2022, 07:36:24 PM

No software would take the private key in (tried a dozen), and trying a WiF decoder showed it's invalid (even though it "looks" right, starts with 5K, right length, etc).
-snip-
Any other ideas of what to do? There's 3 BTC on there.

Let me guess, it's: 5Km2kuu7vtFDPpxywn4u3NLpbr5jKpTB3jsuDU2KYEqetwr388P, right?
I'm sorry to tell you that it's the prvKey FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141 which is out of range, invalid.

Sadly, no one can recover those 3.7 BTC that your "friend" accumulated: 1FYMZEHnszCHKTBdFZ2DLrUuk3dGwYKQxh

Sorry @nc50lc but it does'nt matter it is "out of range".

see:

Code:

import hashlib

g=(0x79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798, 0x483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8)

p = ZZ( '0xFFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFE FFFFFC2F'.replace( ' ', '' ) )

n = ZZ( '0xFFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFE BAAEDCE6 AF48A03B BFD25E8C D0364141'.replace( ' ', '' ) )

E = EllipticCurve(GF(p), [0, 7])

G = E.point( g )

def egcd(a, b):

if a == 0:

return (b, 0, 1)

else:

g, y, x = egcd(b % a, a)

return (g, x - (b // a) * y, y)

def modinv(a, m):

g, x, y = egcd(a, m)

if g != 1:

raise Exception('modular inverse does not exist')

else:

return x % m

def verify(r, s,z,public_key):

w = modinv(s, n)
u1 = (z * w) % n
u2 = (r * w) % n

D=u1*G + u2*public_key

x,y=D.xy()
x=int(x)

if (r % n) == (x % n):
print( "signature matches")

else:
print("invalid signature")

r= 111175281461482630465516451385666215051004681245013976528598462758289754744929
s= 70043377187322970975383334126537096260470471254635274932605589652196963378161
z= 1

x1=65484586321995029360829397682915368247978476961863225607803717802088249892660
y1=72074870721525551148484769172216378998698581912792399280515952501346465251009
P=E.point((x1,y1))
x2=40909554126419277592724504966829837604137845573578049527014144934973709534933
y2=87404510172103350666497040794028294741242353586809580318994867241148928032959
P2=E.point((x2,y2))

verify(r,s,z,P)
verify(r,s,z,P2)

as you see two differents pubkey are valid for the same transactions.

what that means -> need finds "additional" pubkey for valid transactions for addres "0" or "n", then you can spend coins.
realy good mathematician can do.