Bitcoin puzzle transaction ~32 BTC prize to who solves it

[toshisa_toshisa]

🚨🚨 EMERGENCY WARNING FOR BITCOIN PUZZLE #71 HUNTERS 🚨🚨

⚠️ I think anyone involved in Bitcoin Puzzle #71 should be extremely careful with krackpot.io.

Gridinsoft has recently analyzed the website and currently classifies it as:

🚨 MALWARE DISTRIBUTOR
⚠️ TRUST SCORE: 1/100

🔗 Report:
https://gridinsoft.com/online-virus-scanner/url/krackpot-io

According to the report, krackpot.io is advertising itself as:

 “Krackpot: crack Bitcoin Puzzle 71 for 6 BTC”

The site apparently allows users to point their GPU at Bitcoin Puzzle #71 through their browser.

🚩 Some of the warning signals reported by Gridinsoft include:

* ⚠️ Domain is only around 2 months old
* 🚩 Cryptocurrency-related risk indicators
* 🚩 Software/crack-related content indicators
* ⚠️ Low third-party reputation
* ⚠️ Limited independent reputation/history
* 🚩 Suspicious or unclear social-media links
* 🚨 1 external security-provider warning
* 🚨 Gridinsoft currently gives the domain only 1/100 trust

The domain was reportedly registered on July 7, 2026.

⚠️ IMPORTANT: I am NOT saying this proves that the owners are intentionally distributing malware. Gridinsoft itself says its classification is based largely on automated analysis, and automated systems can sometimes produce false positives.

However, considering that Bitcoin Puzzle #71 attracts people running GPU software, downloading tools, entering wallet information, and working with private keys, I think this deserves serious attention.

🚨 DO NOT:

* ❌ Download or execute unknown software from the site
* ❌ Enter private keys
* ❌ Enter seed phrases
* ❌ Connect wallets containing funds
* ❌ Give browser extensions or software unnecessary permissions
* ❌ Disable antivirus/security protections just to make something run

🔐 Until someone from the community independently audits exactly what the website is doing, I would treat it as **HIGH RISK**.

If anybody here has:

* 🔍 Inspected the site's JavaScript
* 🔍 Analyzed its browser GPU/WebGPU code
* 🔍 Captured its network requests
* 🔍 Downloaded and reverse-engineered any files it provides
* 🔍 Actually used the service

please post your findings.

It would be useful to determine whether this is:

⚠️ A genuine security problem
⚠️ A false positive caused by the site's Bitcoin/cracking terminology
⚠️ Or something else entirely

🚨🚨 **PUZZLE #71 HUNTERS — BE CAREFUL.** 🚨🚨

🔐 Never expose your private keys, wallet seeds, or systems holding BTC just because a website promises additional GPU search power.

⚠️ **VERIFY FIRST. TRUST LATER.** ⚠️

[bankeroft]

🚨🚨 EMERGENCY WARNING FOR BITCOIN PUZZLE #71 HUNTERS 🚨🚨

⚠️ I think anyone involved in Bitcoin Puzzle #71 should be extremely careful with krackpot.io.

Gridinsoft has recently analyzed the website and currently classifies it as:

🚨 MALWARE DISTRIBUTOR
⚠️ TRUST SCORE: 1/100

🔗 Report:
https://gridinsoft.com/online-virus-scanner/url/krackpot-io

According to the report, krackpot.io is advertising itself as:

 “Krackpot: crack Bitcoin Puzzle 71 for 6 BTC”

The site apparently allows users to point their GPU at Bitcoin Puzzle #71 through their browser.

🚩 Some of the warning signals reported by Gridinsoft include:

* ⚠️ Domain is only around 2 months old
* 🚩 Cryptocurrency-related risk indicators
* 🚩 Software/crack-related content indicators
* ⚠️ Low third-party reputation
* ⚠️ Limited independent reputation/history
* 🚩 Suspicious or unclear social-media links
* 🚨 1 external security-provider warning
* 🚨 Gridinsoft currently gives the domain only 1/100 trust

The domain was reportedly registered on July 7, 2026.

⚠️ IMPORTANT: I am NOT saying this proves that the owners are intentionally distributing malware. Gridinsoft itself says its classification is based largely on automated analysis, and automated systems can sometimes produce false positives.

However, considering that Bitcoin Puzzle #71 attracts people running GPU software, downloading tools, entering wallet information, and working with private keys, I think this deserves serious attention.

🚨 DO NOT:

* ❌ Download or execute unknown software from the site
* ❌ Enter private keys
* ❌ Enter seed phrases
* ❌ Connect wallets containing funds
* ❌ Give browser extensions or software unnecessary permissions
* ❌ Disable antivirus/security protections just to make something run

🔐 Until someone from the community independently audits exactly what the website is doing, I would treat it as **HIGH RISK**.

If anybody here has:

* 🔍 Inspected the site's JavaScript
* 🔍 Analyzed its browser GPU/WebGPU code
* 🔍 Captured its network requests
* 🔍 Downloaded and reverse-engineered any files it provides
* 🔍 Actually used the service

please post your findings.

It would be useful to determine whether this is:

⚠️ A genuine security problem
⚠️ A false positive caused by the site's Bitcoin/cracking terminology
⚠️ Or something else entirely

🚨🚨 **PUZZLE #71 HUNTERS — BE CAREFUL.** 🚨🚨

🔐 Never expose your private keys, wallet seeds, or systems holding BTC just because a website promises additional GPU search power.

⚠️ **VERIFY FIRST. TRUST LATER.** ⚠️

The rule is that the first person to see the private key will likely claim the prize for themselves and will fail to keep their promises to you. Do not trust any unknown person, website, or program; instead, rely on yourself and your own software.

[toshisa_toshisa]

I found something interesting during my search and wanted to hear what others think.

These are two results:

```
PubAddress: 1PWo3JeYEgBK9dt1egRP3rwMvxvUAnkv17
Priv (HEX): 0x659436EB201A7433FA

PubAddress: 1PWo3JeYEHBw7dfxuPzXB7TGK9jp886EH6
Priv (HEX): 0x659436EB210F344535
```

The private keys are quite close to each other:

```
659436EB201A7433FA
659436EB210F344535
```

Difference:

```
0xF4C0113B
= 4,106,228,027
```

What caught my attention is the addresses.

Character-by-character:

```
1PWo3JeYEgBK9dt1egRP3rwMvxvUAnkv17
1PWo3JeYEHBw7dfxuPzXB7TGK9jp886EH6
  ||||||||| |  |
```

They have the same characters at **11 out of 34 positions**.

Matching positions:

```
1  = 1
2  = P
3  = W
4  = o
5  = 3
6  = J
7  = e
8  = Y
9  = E
11 = B
14 = d
```

So both addresses start with exactly:

```
1PWo3JeYE
```

and they also match again at positions 11 and 14.

The important part is that I understand nearby private keys should NOT normally generate visually similar Bitcoin addresses. Because of EC public-key generation followed by SHA-256 and RIPEMD-160, small changes in the private key should effectively produce unrelated address hashes.

So my question is:

**Is this simply an interesting statistical coincidence, or is there anything worth investigating when two relatively close private keys generate addresses sharing this many characters in exactly the same positions?**

I'm not claiming there is a correlation. I'm interested in the probability/statistics behind seeing a pair like this.

Especially interested to hear from anyone who has done large-scale VanitySearch/KeyHunt scans and has seen similar clusters.

[bankeroft]

I found something interesting during my search and wanted to hear what others think.

These are two results:

```
PubAddress: 1PWo3JeYEgBK9dt1egRP3rwMvxvUAnkv17
Priv (HEX): 0x659436EB201A7433FA

PubAddress: 1PWo3JeYEHBw7dfxuPzXB7TGK9jp886EH6
Priv (HEX): 0x659436EB210F344535
```

The private keys are quite close to each other:

```
659436EB201A7433FA
659436EB210F344535
```

Difference:

```
0xF4C0113B
= 4,106,228,027
```

What caught my attention is the addresses.

Character-by-character:

```
1PWo3JeYEgBK9dt1egRP3rwMvxvUAnkv17
1PWo3JeYEHBw7dfxuPzXB7TGK9jp886EH6
  ||||||||| |  |
```

They have the same characters at **11 out of 34 positions**.

Matching positions:

```
1  = 1
2  = P
3  = W
4  = o
5  = 3
6  = J
7  = e
8  = Y
9  = E
11 = B
14 = d
```

So both addresses start with exactly:

```
1PWo3JeYE
```

and they also match again at positions 11 and 14.

The important part is that I understand nearby private keys should NOT normally generate visually similar Bitcoin addresses. Because of EC public-key generation followed by SHA-256 and RIPEMD-160, small changes in the private key should effectively produce unrelated address hashes.

So my question is:

**Is this simply an interesting statistical coincidence, or is there anything worth investigating when two relatively close private keys generate addresses sharing this many characters in exactly the same positions?**

I'm not claiming there is a correlation. I'm interested in the probability/statistics behind seeing a pair like this.

Especially interested to hear from anyone who has done large-scale VanitySearch/KeyHunt scans and has seen similar clusters.

rare statistical coincidence !!!

[detechs]

I found something interesting during my search and wanted to hear what others think.

These are two results:

```
PubAddress: 1PWo3JeYEgBK9dt1egRP3rwMvxvUAnkv17
Priv (HEX): 0x659436EB201A7433FA

PubAddress: 1PWo3JeYEHBw7dfxuPzXB7TGK9jp886EH6
Priv (HEX): 0x659436EB210F344535
```

The private keys are quite close to each other:

```
659436EB201A7433FA
659436EB210F344535
```

Difference:

```
0xF4C0113B
= 4,106,228,027
```

What caught my attention is the addresses.

Character-by-character:

```
1PWo3JeYEgBK9dt1egRP3rwMvxvUAnkv17
1PWo3JeYEHBw7dfxuPzXB7TGK9jp886EH6
  ||||||||| |  |
```

They have the same characters at **11 out of 34 positions**.

Matching positions:

```
1  = 1
2  = P
3  = W
4  = o
5  = 3
6  = J
7  = e
8  = Y
9  = E
11 = B
14 = d
```

So both addresses start with exactly:

```
1PWo3JeYE
```

and they also match again at positions 11 and 14.

The important part is that I understand nearby private keys should NOT normally generate visually similar Bitcoin addresses. Because of EC public-key generation followed by SHA-256 and RIPEMD-160, small changes in the private key should effectively produce unrelated address hashes.

So my question is:

**Is this simply an interesting statistical coincidence, or is there anything worth investigating when two relatively close private keys generate addresses sharing this many characters in exactly the same positions?**

I'm not claiming there is a correlation. I'm interested in the probability/statistics behind seeing a pair like this.

Especially interested to hear from anyone who has done large-scale VanitySearch/KeyHunt scans and has seen similar clusters.

Keep looking into it, cause I definitely am, anyone saying no pattern or coincidence is probably hoping u don't investigate it

[eggsylacer]

Did anyone manage to arrange the points on the curve in their own order?
Is this even possible? As far as I understand, something similar is implemented in the “kangaroo” algorithm.

[optioncmdPR]

It is possible. Regarding implementing and testing the results of customized elliptic curve parameters, you can play around with this:

Code:

P = 13407807929942597099574024998205846127479365820592393377723561443721764030073431184712636981971479856705023170278632780869088242247907112362425735876442206
  # 2^512 - 2^256 - 1954
Gx =0xf37cccfdf3b97758ab40c52b9d0e160e0537f9b65b9c51b2b3e502b62df02f30
Gy =0x9075b4ee4d4788cabb49f7f81c221151fa2f68914d0aa833388fa11ff621a970

def mod(a, b=P):
    res = a % b
    return res if res >= 0 else res + b

def mod_inverse(a, m=P):
    b, u, v = m, 0, 1
    old_a, old_u = a, 1
    while old_a != 0:
        q = b // old_a
        b, old_a = old_a, b - q * old_a
        u, old_u = old_u, u - q * old_u
    return mod(u, m)

def point_add(p1, p2):
    if not p1: return p2
    if not p2: return p1
    
    if p1['x'] == p2['x']:
        if p1['y'] != p2['y'] or p1['y'] == 0: 
            return None
        m = mod(3 * p1['x'] * p1['x'] * mod_inverse(2 * p1['y']))
        x = mod(m * m - 2 * p1['x'])
        y = mod(m * (p1['x'] - x) - p1['y'])
        return {'x': x, 'y': y}
    else:
        m = mod(mod(p2['y'] - p1['y']) * mod_inverse(mod(p2['x'] - p1['x'])))
        x = mod(m * m - p1['x'] - p2['x'])
        y = mod(m * (p1['x'] - x) - p1['y'])
        return {'x': x, 'y': y}

def point_multiply(k):
    if not k or k <= 0: 
        k = 1
    result = None
    addend = {'x': Gx, 'y': Gy}
    scalar = k
    while scalar > 0:
        if scalar & 1:
            result = point_add(result, addend)
        addend = point_add(addend, addend)
        scalar >>= 1
    return result or {'x': Gx, 'y': Gy}

def to_base_n(num: int, base: int) -> str:
    """Converts a large integer to a string representation in any base (2 to 10)."""
    if num == 0:
        return "0"
    digits = []
    while num > 0:
        digits.append(str(num % base))
        num //= base
    return "".join(reversed(digits))



if __name__ == "__main__":
    print("=== Elliptic Curve Scalar Multiplication ===")
    print("Supports Decimal (e.g., 5), Hex (0x05), or Binary (0b101)\n")
    
    try:
        # Prompt for input
        user_input = input("Enter scalar value: ").strip()
        
        # Determine format and parse dynamically
        if user_input.lower().startswith('0x'):
            scalar = int(user_input, 16)
        elif user_input.lower().startswith('0b'):
            scalar = int(user_input, 2)
        else:
            scalar = int(user_input)
            
        print("\nCalculating point multiplication...")
        public_key = point_multiply(scalar)
        
        x_val = public_key['x']
        y_val = public_key['y']
        
        # Display Results
        print("\n" + "="*60)
        print(f"Input Scalar (Decimal): {scalar}")
        print(f"Input Scalar (Hex):     {hex(scalar)}")
        print("="*60)
        
        print("\n--- POINT X COORDINATE ---")
        print(f"Base 10 : {x_val}")
        print(f"Base 9  : {to_base_n(x_val, 9)}")
        print(f"Base 5  : {to_base_n(x_val, 5)}")
        
        print("\n--- POINT Y COORDINATE ---")
        print(f"Base 10 : {y_val}")
        print(f"Base 9  : {to_base_n(y_val, 9)}")
        print(f"Base 5  : {to_base_n(y_val, 5)}")
        print("="*60)

    except ValueError:
        print("\n[Error] Invalid input format. Please check your value.")
    except Exception as e:
        print(f"\n[Error] An unexpected error occurred: {e}")

    # Final pause ensures the terminal window stays open
    input("\nProcess finished. Press Enter to close this window...")

[fecell]

hi, buddy!

how much points U can check at one sec with own programs?
i'v got 115m/s (at cold start) to 104-108m/s at work state with 1650ti (4gb, notebook). 
for 140bits.

whats 'bout Ur algo?

Code:

[15:51:34] [+] [C 21,706 / 2,596,148,429,267,413,814,265,248,164,610,048] [11,653,320,015,872 / 696,898,287,454,081,973,172,991,196,020,261,297,061,888 prob] S: 104591380.9 pr/s, E: 1d 06:56:57, ETA: 3.390e+26 y, exp-ETA: 1.353e+17 y, progress: 0.0000%

[fecell]

It is possible. Regarding implementing and testing the results of customized elliptic curve parameters, you can play around with this:

sure. it's like child playing game.!

[eggsylacer]

It is possible. Regarding implementing and testing the results of customized elliptic curve parameters, you can play around with this:

Code:

P = 13407807929942597099574024998205846127479365820592393377723561443721764030073431184712636981971479856705023170278632780869088242247907112362425735876442206
  # 2^512 - 2^256 - 1954
Gx =0xf37cccfdf3b97758ab40c52b9d0e160e0537f9b65b9c51b2b3e502b62df02f30
Gy =0x9075b4ee4d4788cabb49f7f81c221151fa2f68914d0aa833388fa11ff621a970

def mod(a, b=P):
    res = a % b
    return res if res >= 0 else res + b

def mod_inverse(a, m=P):
    b, u, v = m, 0, 1
    old_a, old_u = a, 1
    while old_a != 0:
        q = b // old_a
        b, old_a = old_a, b - q * old_a
        u, old_u = old_u, u - q * old_u
    return mod(u, m)

def point_add(p1, p2):
    if not p1: return p2
    if not p2: return p1
    
    if p1['x'] == p2['x']:
        if p1['y'] != p2['y'] or p1['y'] == 0: 
            return None
        m = mod(3 * p1['x'] * p1['x'] * mod_inverse(2 * p1['y']))
        x = mod(m * m - 2 * p1['x'])
        y = mod(m * (p1['x'] - x) - p1['y'])
        return {'x': x, 'y': y}
    else:
        m = mod(mod(p2['y'] - p1['y']) * mod_inverse(mod(p2['x'] - p1['x'])))
        x = mod(m * m - p1['x'] - p2['x'])
        y = mod(m * (p1['x'] - x) - p1['y'])
        return {'x': x, 'y': y}

def point_multiply(k):
    if not k or k <= 0: 
        k = 1
    result = None
    addend = {'x': Gx, 'y': Gy}
    scalar = k
    while scalar > 0:
        if scalar & 1:
            result = point_add(result, addend)
        addend = point_add(addend, addend)
        scalar >>= 1
    return result or {'x': Gx, 'y': Gy}

def to_base_n(num: int, base: int) -> str:
    """Converts a large integer to a string representation in any base (2 to 10)."""
    if num == 0:
        return "0"
    digits = []
    while num > 0:
        digits.append(str(num % base))
        num //= base
    return "".join(reversed(digits))



if __name__ == "__main__":
    print("=== Elliptic Curve Scalar Multiplication ===")
    print("Supports Decimal (e.g., 5), Hex (0x05), or Binary (0b101)\n")
    
    try:
        # Prompt for input
        user_input = input("Enter scalar value: ").strip()
        
        # Determine format and parse dynamically
        if user_input.lower().startswith('0x'):
            scalar = int(user_input, 16)
        elif user_input.lower().startswith('0b'):
            scalar = int(user_input, 2)
        else:
            scalar = int(user_input)
            
        print("\nCalculating point multiplication...")
        public_key = point_multiply(scalar)
        
        x_val = public_key['x']
        y_val = public_key['y']
        
        # Display Results
        print("\n" + "="*60)
        print(f"Input Scalar (Decimal): {scalar}")
        print(f"Input Scalar (Hex):     {hex(scalar)}")
        print("="*60)
        
        print("\n--- POINT X COORDINATE ---")
        print(f"Base 10 : {x_val}")
        print(f"Base 9  : {to_base_n(x_val, 9)}")
        print(f"Base 5  : {to_base_n(x_val, 5)}")
        
        print("\n--- POINT Y COORDINATE ---")
        print(f"Base 10 : {y_val}")
        print(f"Base 9  : {to_base_n(y_val, 9)}")
        print(f"Base 5  : {to_base_n(y_val, 5)}")
        print("="*60)

    except ValueError:
        print("\n[Error] Invalid input format. Please check your value.")
    except Exception as e:
        print(f"\n[Error] An unexpected error occurred: {e}")

    # Final pause ensures the terminal window stays open
    input("\nProcess finished. Press Enter to close this window...")

Buddy, I don't need an intermediary between me and the AI. No hard feelings.
Moreover, the question was deliberately constructed in such a way that the AI couldn’t understand it, while someone who truly understands the topic could.