Ask people I don't know on Telegram what kind of personality I am.
My aim is to provide full support and unity.
Sometimes very absurd ideas are discussed. But I don't make fun of them. Because HUMAN is a being that is open to improving itself every day.
Yes, I have strict rules. If there is no RESPECT, there is no community. You cannot create a community with a person who is DISRESPECTFUL.


A few notes,

- I think WP 71. stopped looking for wallets.

We had good memories with WP. Sometimes we argued.. We even did a small test with him. He found 1 1PWo3JeB9j prefix.

- We had a conversation about this with a few people I don't know. Did they have a problem with TRUST? (You can ask.)

- I have made a few sincere friends now. Because I didn't write anything wrong to anyone. I didn't say anything bad. I RESPECTED them.

- It's nice to argue or agree with someone. But if it goes beyond the level of argument, DISRESPECT begins. I react to this. (Because don't ask people who don't believe about their beliefs. You'll waste your time.)

Aren't you? An old boy trying to apologize. lol

Are you talking to me? Have you improved yourself that much? ) Were you using artificial intelligence 5.0?

Now your lover is waiting. (GPT) Don't answer me without asking him.

Attention other friends.!!

Your knowledge about prefixes is really weak. I recommend you to do research on how to improve this.
To those who don't believe my idea here, I say HODRİ MEYDAN.
Does anyone have more and closer h160s than me?
I know the answer, no.

I am writing the ones that are close for example. Compare your own h160.

f6f5431d25bbf7ebf2d652ebdd4706567468968b
f6f5431d25bbf7c73fc6ba422e2c9384944e01e5
f6f5431d25bbf7c2c1a542f87f2b4c6fcd994ce9
f6f5431d25bbf7383f07a040f6089a59d4d725e2
f6f5431d25bbf71fe311d2d4b68fe67f315466a8
f6f5431d25bbf702bafee0413954bfe84354863a
...

Public Addr: 1PWo3JeBbURW5W5W6exwvLHfvRMRQ28SP7
Priv (WIF): p2pkh:KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3rKL1DAnemGxU2wWDbE7V
Priv (HEX): 0x                                              794B37D7B6E94E4F5F  

Public Addr: 1PWo3JeBWCXEVkzycsVwm5bs3d3aj2J9hB
Priv (WIF): p2pkh:KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3rKL1DBLVYzWiZu3f2zSc
Priv (HEX): 0x                                              794B37D9EDA9E85057   

Public Addr: 1PWo3JeBeJfj6GY3Uk2zX6ZCMxJtPNsGTF
Priv (WIF): p2pkh:KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3rKL1DBTAE2PrM7wGtNtM
Priv (HEX): 0x                                              794B37DA6453F75BF3   

Public Addr: 1PWo3JeBergQXkCKD1gTN29HZXRTgEJw8h
Priv (WIF): p2pkh:KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3rKL1DBaxzNYdiPxBUSxr
Priv (HEX): 0x                                              794B37DAEF465C322C

I noticed that some people in the group are searching for prefixes of the Bitcoin address:

1PWo3JeB9jrGwfHDNpdGK54CRas7fsVzXU

For example: 1PWo3JeB, and so on.
Can anyone with experience explain the reasoning behind this?

It doesn’t make sense to me, since the private key is what gets incremented, and each key generates a completely different Bitcoin address.

What am I missing here?!!!

Where i can get the RCKangaro exe for GPU search

If you want in my GitHub you can find VanitySearch-bitcrack repo, 50% faster than these

If you suspect that there is a backdoor in it, you can always use it offline.
If it doesn not work offline, than you should be worried about it.
But I don't think ICELAND is puting some backdoors in his apps...

hanks so much for the detailed breakdown! 🙏

I’ve got a basic grasp of modular arithmetic, but initially interpreted the situation differently — like, the point −𝑃 −P kinda feels “greater” than 𝑃
P, especially when you look at it through the lens of inversion over the 𝑦 y-axis. 📉📈

Now here’s the spicy bit 🌶️:
If you start walking from − 𝑃 −P, incrementing the private key by 1 each time (i.e., checking for some  𝑘 k where 𝑘⋅𝐺=𝑃 k⋅G=P), you'll actually hit the point 𝑃 P relatively quickly — because they’re connected through the curve’s group order.
But if you go the other way — starting at 𝑃
P, trying to reach −𝑃
−P via +1 steps — you'll have to walk almost the entire group order. 🚶‍♂️🔁

That means direction matters, and that’s something we might be able to leverage. This method gives a hint about the sign or orientation of the point. 👀🧭

🤔 I'm wondering — are there any libs or implementations that let you determine which of two points is “closer to 0” or gives clues about their directional relationship?

💡 Also, are there any side channels we can sniff — like timing differences in signature verification, subtle quirks in point serialization, byte structure anomalies, or even differences when decompressing points — that could leak info about the point’s sign or help narrow the keyspace?

If there’s even the faintest trace of something like that — it’s a potential vulnerability. Would love to hear thoughts from anyone who’s dug deep into libsecp256k1 or forks. 🔬🔓

Again — massive thanks for the insights. Really appreciate the depth! 🚀🧠💥

I this  code goes get ready beat rc coder

There was need to modify the code. Works fine now.

How to edit VanitySearch by JLP to search in puzzle 71 space (not from 1)?

Anybody gone into c/c++ in this project?

why are you cluttering this thread with your useless pompous comments, stop hijacking peoples threads

OP asked to post speed results of different cards, so I did, it was my first post ever on this forum, and I purposely
 joined so that I can post my results and try to contribute some data to this thread, and then you reply to it with your snotty comment, what is wrong with you, get over yourself, you are not that important


Just because I showed interest in his hardware setup of 400 gpus, does not mean I want to be spoon fed, I was just wanting to see what a 400 gpu setup looks like.

I did look threw his post history and did not see pictures, maybe he posted them some where else, did you happen do consider that?

from ecdsa.ellipticcurve import Point
from ecdsa.curves import SECP256k1

# Secp256k1 parameters
curve = SECP256k1.curve
p = curve.p()
n = SECP256k1.order
G = SECP256k1.generator

# Endomorphism constants
beta = 0x7ae96a2b657c07106e64479eac3434e99cf0497512f58995c1396c28719501ee
lmbda = 0x5363ad4cc05c30e0a5261c028812645a122e22ea20816678df02967c1b23bd72
beta2 = (beta * beta) % p
lmbda2 = (lmbda * lmbda) % n

def endomorphism(P, beta_val):
    """Apply x-coordinate endomorphism with given beta"""
    return Point(curve, (beta_val * P.x()) % p, P.y(), order=n)

def negate_point(P):
    """Negate point by flipping y-coordinate"""
    return Point(curve, P.x(), (-P.y()) % p, order=n)

def format_key(P):
    """Format point as uncompressed public key"""
    return f'04{P.x():064x}{P.y():064x}'

# Input private key
k = 5

# Base point
P = k * G

# Generate all 6 keys
keys = [
    # Original y group
    (P, k),
    (endomorphism(P, beta), (k * lmbda) % n),
    (endomorphism(P, beta2), (k * lmbda2) % n),
   
    # Negated y group
    (negate_point(P), (n - k) % n),
    (negate_point(endomorphism(P, beta)), (n - (k * lmbda)) % n),
    (negate_point(endomorphism(P, beta2)), (n - (k * lmbda2)) % n)
]

# Print results
print(f"Uncompressed public key: {format_key(P)}\n")

print("Three x values:")
print(f"x1 = {P.x()}")
print(f"x2 = {endomorphism(P, beta).x()}")
print(f"x3 = {endomorphism(P, beta2).x()}\n")

print("Two y values:")
print(f"y1 = {P.y()}")
print(f"y2 = {negate_point(P).y()}\n")

print("Six public keys with private keys:")
for i, (point, priv) in enumerate(keys, 1):
    print(f"Public key {i}: {format_key(point)} [Private key: {priv}]")
    print(f"  Validated: {point == priv * G}")

sum_y1 = sum(keys[1] for i in range(3))
sum_y2 = sum(keys[1] for i in range(3,6))

print(f"\nSum for y1: {sum_y1} (n = {n})")
print(f"Sum for y2: {sum_y2} (2n = {2*n})")

from ecdsa.ellipticcurve import Point
from ecdsa.curves import SECP256k1

# Secp256k1 parameters
curve = SECP256k1.curve
p = curve.p()
n = SECP256k1.order
G = SECP256k1.generator

# Endomorphism constants
beta = 0x7ae96a2b657c07106e64479eac3434e99cf0497512f58995c1396c28719501ee
lmbda = 0x5363ad4cc05c30e0a5261c028812645a122e22ea20816678df02967c1b23bd72
beta2 = (beta * beta) % p
lmbda2 = (lmbda * lmbda) % n

def endomorphism(P, beta_val):
    """Apply x-coordinate endomorphism with given beta"""
    return Point(curve, (beta_val * P.x()) % p, P.y(), order=n)

def negate_point(P):
    """Negate point by flipping y-coordinate"""
    return Point(curve, P.x(), (-P.y()) % p, order=n)

def format_key(P):
    """Format point as uncompressed public key"""
    return f'04{P.x():064x}{P.y():064x}'

# Input private key
k = 5

# Base point
P = k * G

# Generate all 6 keys
keys = [
    # Original y group
    (P, k),
    (endomorphism(P, beta), (k * lmbda) % n),
    (endomorphism(P, beta2), (k * lmbda2) % n),
   
    # Negated y group
    (negate_point(P), (n - k) % n),
    (negate_point(endomorphism(P, beta)), (n - (k * lmbda)) % n),
    (negate_point(endomorphism(P, beta2)), (n - (k * lmbda2)) % n)
]

# Print results
print(f"Uncompressed public key: {format_key(P)}\n")

print("Three x values:")
print(f"x1 = {P.x()}")
print(f"x2 = {endomorphism(P, beta).x()}")
print(f"x3 = {endomorphism(P, beta2).x()}\n")

print("Two y values:")
print(f"y1 = {P.y()}")
print(f"y2 = {negate_point(P).y()}\n")

print("Six public keys with private keys:")
for i, (point, priv) in enumerate(keys, 1):
    print(f"Public key {i}: {format_key(point)} [Private key: {priv}]")
    print(f"  Validated: {point == priv * G}")

sum_y1 = sum(keys[1] for i in range(3))
sum_y2 = sum(keys[1] for i in range(3,6))

print(f"\nSum for y1: {sum_y1} (n = {n})")
print(f"Sum for y2: {sum_y2} (2n = {2*n})")

I learn many things from this smaller secp256k1.

Uncompressed public key: 042f8bde4d1a07209355b4a7250a5c5128e88b84bddc619ab7cba8d569b240efe4d8ac222636e5e 3d6d4dba9dda6c9c426f788271bab0d6840dca87d3aa6ac62d6

Three x values:
x1 = 21505829891763648114329055987619236494102133314575206970830385799158076338148
x2 = 23285849548026170226712523888619559634478006467037872208296602441247713932904
x3 = 71000409797526377082529405132449111724689844884027484860330595767503044400611

Two y values:
y1 = 98003708678762621233683240503080860129026887322874138805529884920309963580118
y2 = 17788380558553574189887744505607047724243097342766425233927699087598871091545
Enter private key for (x1, y1) = (21505829891763648114329055987619236494102133314575206970830385799158076338148, 98003708678762621233683240503080860129026887322874138805529884920309963580118): 5

Six public keys with private keys:
Public key 1: 042f8bde4d1a07209355b4a7250a5c5128e88b84bddc619ab7cba8d569b240efe4d8ac222636e5e 3d6d4dba9dda6c9c426f788271bab0d6840dca87d3aa6ac62d6 [Private key: 5]
  Validated: k1 matches
Public key 2: 04337b52e3acda49dff79f54fbccb94671a045693ee0d097cc138c694695a83668d8ac222636e5e 3d6d4dba9dda6c9c426f788271bab0d6840dca87d3aa6ac62d6 [Private key: 72798312578463789091060122408687194401800723498338030560477939572921828405753]
  Validated: k2 matches
Public key 3: 049cf8cecf391e958cb2ac03df28ea6865772f120342cdcd7c20cac14eb816d5e3d8ac222636e5e 3d6d4dba9dda6c9c426f788271bab0d6840dca87d3aa6ac62d6 [Private key: 42993776658852406332510862600000713451036840780736873822127223568596333088579]
  Validated: k3 matches
Public key 4: 042f8bde4d1a07209355b4a7250a5c5128e88b84bddc619ab7cba8d569b240efe42753ddd9c91a1 c292b24562259363bd90877d8e454f297bf235782c459539959 [Private key: 115792089237316195423570985008687907852837564279074904382605163141518161494332]
  Validated: k4 matches
Public key 5: 04337b52e3acda49dff79f54fbccb94671a045693ee0d097cc138c694695a836682753ddd9c91a1 c292b24562259363bd90877d8e454f297bf235782c459539959 [Private key: 42993776658852406332510862600000713451036840780736873822127223568596333088584]
  Validated: k5 matches
Public key 6: 049cf8cecf391e958cb2ac03df28ea6865772f120342cdcd7c20cac14eb816d5e32753ddd9c91a1 c292b24562259363bd90877d8e454f297bf235782c459539959 [Private key: 72798312578463789091060122408687194401800723498338030560477939572921828405758]
  Validated: k6 matches

Sum of private keys for y1 = 115792089237316195423570985008687907852837564279074904382605163141518161494337 (should be n or 2n)
Sum of private keys for y2 = 231584178474632390847141970017375815705675128558149808765210326283036322988674 (should be n or 2n)

The 2 people you mentioned,
RetirecCoder = The only individual with only money power, someone who does not have any software that reaches different speeds. He broke it ONLY with his MONEY POWER.