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    Author Topic: Mini-puzzle #4  (Read 946 times)
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    • RetiredCoder (OP)
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    January 14, 2025, 05:22:30 PM
     #1
    Guys, let's have some fun again, I have one more mini-puzzle for you Smiley
    There is 0.01 BTC on that address, so hurry up!

    03C8F139DAD58B4786D3649992849733C2A7626F011089E87508CCDF8B0758C493
    Hint: It's a strong point so you cannot solve it by kangaroos. But it's a special point, you can solve it easily if you use some feature of secp256k1.

    PS. No BS here please, I will remove it.
    PPS. For history, previous mini-puzzle is here: https://bt.irlbtc.com/view/5522785
    I've solved #120, #125, #130. How: https://github.com/RetiredC

  • llenn1227
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    January 14, 2025, 05:23:52 PM
     #2
    Quote from: RetiredCoder on January 14, 2025, 05:22:30 PM
    Guys, let's have some fun again, I have one more mini-puzzle for you Smiley
    There is 0.01 BTC on that address, so hurry up!

    03C8F139DAD58B4786D3649992849733C2A7626F011089E87508CCDF8B0758C493
    Hint: It's a strong point so you cannot solve it by kangaroos. But it's a special point, you can solve it easily if you use some feature of secp256k1.

    PS. No BS here please, I will remove it.
    PPS. For history, previous mini-puzzle is here: https://bt.irlbtc.com/view/5522785



    That's Sad, a hard one QAQ

    But i'll try my best Cheesy

    Here's the address - > https://www.blockchain.com/explorer/addresses/btc/1AH5pRZW4ZofEJhdb3muNZ1q88YNu3Rez7

  • RetiredCoder (OP)
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    January 14, 2025, 05:26:13 PM
     #3
    Quote from: llenn1227 on January 14, 2025, 05:23:52 PM
    That's Sad, a hard one QAQ
    But i'll try my best Cheesy

    It's not difficult, really  Wink
    And you don't even need GPUs or fast PC.
    I've solved #120, #125, #130. How: https://github.com/RetiredC

  • llenn1227
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    January 14, 2025, 05:30:19 PM
    Last edit: January 14, 2025, 06:31:28 PM by llenn1227
     #4
    Quote from: RetiredCoder on January 14, 2025, 05:26:13 PM
    Quote from: llenn1227 on January 14, 2025, 05:23:52 PM
    That's Sad, a hard one QAQ
    But i'll try my best Cheesy

    It's not difficult, really  Wink
    And you don't even need GPUs or fast PC.


    I think i need a better brain.
    For a pool guy who failed 4 subject in this semester.
    Maybe I need to study whole ECDSA for the next puzzle?

  • RetiredCoder (OP)
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    January 14, 2025, 06:58:05 PM
     #5
    Quote from: llenn1227 on January 14, 2025, 05:30:19 PM
    Maybe I need to study whole ECDSA for the next puzzle?

    ECDSA already was in previous puzzle, so probably I won't use it again  Smiley
    I've solved #120, #125, #130. How: https://github.com/RetiredC

  • llenn1227
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    January 14, 2025, 07:02:25 PM
     #6
    Quote from: RetiredCoder on January 14, 2025, 06:58:05 PM
    Quote from: llenn1227 on January 14, 2025, 05:30:19 PM
    Maybe I need to study whole ECDSA for the next puzzle?

    ECDSA already was in previous puzzle, so probably I won't use it again  Smiley

    That a good news!
    But first i need to figure this puzzle out.
    I guess this might have something to do with the mirrored curve?
    Or maybe the constant of secp256k1?
    Still learning how this work Orz

  • iceland2k14
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    January 15, 2025, 05:30:51 AM
     #7
    Quote from: RetiredCoder on January 14, 2025, 05:26:13 PM

    It's not difficult, really  Wink
    And you don't even need GPUs or fast PC.

    • None of the Endo/Sym points are below 53 bits
    • None of the Endo/Sym points are within 30 bits of the bit boundary from 1 to 256
    • None of the Endo/Sym points are a hash of "Mini-puzzle #4" or "RetiredCoder Mini-puzzle #4"
    • None of the Endo/Sym points are a close to the privatekey or K or previous #3 Puzzle
    • C'mon Guys. do it fast. What else...

  • RetiredCoder (OP)
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    January 15, 2025, 06:15:12 AM
     #8
    No worries, I will give a new hint after 24 hours.
    But I still think that you will solve it earlier Smiley
    I've solved #120, #125, #130. How: https://github.com/RetiredC

  • hskun
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    January 15, 2025, 09:02:56 AM
     #9
    not the p n endo lambda beta...
    it was solved...so sad...

  • ukn9977
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    January 15, 2025, 09:44:00 AM
    Merited by RetiredCoder (1)
     #10
    Thank you @RetiredCoder for the puzzle.

    Because: k = k*LAMBDA+1
    Solve    : k = 89767419612596129250022730500283612052200928704544131917942315608641440322898

    Code:
    # sagemath
    N = 115792089237316195423570985008687907852837564279074904382605163141518161494337
    LAMBDA = 37718080363155996902926221483475020450927657555482586988616620542887997980018

    R.<k> = PolynomialRing(GF(N))

    # k = k*LAMBDA+1
    r = (k -  (k*LAMBDA+1)).roots()
    for x in r:
        print("k = ", x[0])
    #k =  89767419612596129250022730500283612052200928704544131917942315608641440322898

  • RetiredCoder (OP)
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    January 15, 2025, 10:27:05 AM
     #11
    Another explanation:
    It's a special point, if you apply endomorphism for it, you will get previous point (Point - G).
    It's funny to have two points next to each other with same Y, isn't it?  Smiley
    And since you know the distance between these two points, you can calculate the private key easily.
    Congrats to the winner! Smiley
    I've solved #120, #125, #130. How: https://github.com/RetiredC

  • JDScreesh
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    January 15, 2025, 10:54:56 AM
     #12
    Hello there  Smiley

    Congratulations to the winner  Wink

    I think I'm lost with this red part of the explanation:

    Quote from: RetiredCoder on January 15, 2025, 10:27:05 AM
    Another explanation:
    It's a special point, if you apply endomorphism for it, you will get previous point (Point - G).
    It's funny to have two points next to each other with same Y, isn't it?  Smiley
    And since you know the distance between these two points, you can calculate the private key easily.
    Congrats to the winner! Smiley



    I'm trying to get the privkey with this information, but I think I'm missing something  Huh

    Thanks for all the learning  Smiley

    Greetings.
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  • Major Azul
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    January 15, 2025, 11:00:41 AM
     #13
    hello, where can I learn how to decode this?

    and what is the address?

  • RetiredCoder (OP)
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    January 15, 2025, 11:06:49 AM
     #14
    Quote from: JDScreesh on January 15, 2025, 10:54:56 AM
    I'm trying to get the privkey with this information, but I think I'm missing something  Huh
    Thanks for all the learning  Smiley
    Greetings.

    Sure!  Cheesy
    We know:
    lambda * x = a [mod N]
    x = a + 1 [mod N]

    So:
    lambda * (a + 1) = a [mod N]
    lambda * a + lambda = a [mod N]
    (lambda - 1) * a + lambda = 0 [mod N]
    a = -lambda (lambda - 1) [mod N]
    I've solved #120, #125, #130. How: https://github.com/RetiredC

  • JDScreesh
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    January 15, 2025, 11:19:18 AM
     #15
    Quote from: RetiredCoder on January 15, 2025, 11:06:49 AM
    Sure!  Cheesy
    We know:
    lambda * x = a [mod N]
    x = a + 1 [mod N]

    So:
    lambda * (a + 1) = a [mod N]
    lambda * a + lambda = a [mod N]
    (lambda - 1) * a + lambda = 0 [mod N]
    a = -lambda (lambda - 1) [mod N]

    Cool! now I can see it  Grin

    Thanks a lot for the explanation  Wink
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    LaAHGMjisi2NFMfHJpMywKm8ALzmivdRaq

  • mjojo
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    January 15, 2025, 11:34:12 AM
     #16
    Code:
    DBL 0:
    a0                               : 28656494787849788178846280232722535366602610135464471548693628842033875030774 Y
    a1                               : 64085527874626556120582535604906191133993776464141088401876389432686378436844 N
    Lam Numerator                    : 24451327812764160469375732394646891906403158240170826054096020762142281864872 Y
    Lam Denominator                  : 12378966511936916817594086201124474414717568262641612764295194857463922202025 Y
    Result Lam                       : 90164811066602240685525910245268168444406468821116643963079939878016814643363 N
    Modular Inverse                  : 73354833296120792775670026364214776218043641652330273932372744839188764773793 Y
    Result X                         : 90888779158563823146755473862282102363969308959424961308149641532476281635987 Y
    Result Y                         : 46201727910540285190454324890558278619756708117972953873730558795650959119923 N



    DBL 1:
    a0                               : 104403551840626474953376424551165765467561720381305370579875450682007898015831 Y
    a1                               : 64085527874626556120582535604906191133993776464141088401876389432686378436844 N
    Lam Numerator                    : 87957283359521357687061983675423052553526169065692381021112082220578186202170 N
    Lam Denominator                  : 12378966511936916817594086201124474414717568262641612764295194857463922202025 Y
    Result Lam                       : 94165601948352868544180331085494877662307814315103203626908763534409097566489 N
    Modular Inverse                  : 73354833296120792775670026364214776218043641652330273932372744839188764773793 Y
    Result X                         : 23620854997921501804136411791104769986833072451863613446092941013256551136065 Y
    Result Y                         : 46201727910540285190454324890558278619756708117972953873730558795650959119923 N

    same Y coordinate from two point, but still don't know how to calculate Lam 37718080363155996902926221483475020450927657555482586988616620542887997980018

  • JDScreesh
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    January 15, 2025, 11:48:54 AM
     #17
    Quote from: mjojo on January 15, 2025, 11:34:12 AM

    same Y coordinate from two point, but still don't know how to calculate Lam 37718080363155996902926221483475020450927657555482586988616620542887997980018

    I got lambda value searching... and found it in Hal Finney's post "Speeding up signature verification" ( https://bt.irlbtc.com/view/3238.0 )

    The lambda and beta values in that post are in hexadecimal.

    lambda = 0x5363ad4cc05c30e0a5261c028812645a122e22ea20816678df02967c1b23bd72
    beta = 0x7ae96a2b657c07106e64479eac3434e99cf0497512f58995c1396c28719501ee

     Smiley
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    LaAHGMjisi2NFMfHJpMywKm8ALzmivdRaq

  • mjojo
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    January 15, 2025, 12:01:50 PM
     #18
    Quote from: JDScreesh on January 15, 2025, 11:48:54 AM
    Quote from: mjojo on January 15, 2025, 11:34:12 AM

    same Y coordinate from two point, but still don't know how to calculate Lam 37718080363155996902926221483475020450927657555482586988616620542887997980018

    I got lambda value searching... and found it in Hal Finney's post "Speeding up signature verification" ( https://bt.irlbtc.com/view/3238.0 )

    The lambda and beta values in that post are in hexadecimal.

    lambda = 0x5363ad4cc05c30e0a5261c028812645a122e22ea20816678df02967c1b23bd72
    beta = 0x7ae96a2b657c07106e64479eac3434e99cf0497512f58995c1396c28719501ee

     Smiley

    thank you

  • iceland2k14
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    January 15, 2025, 12:09:39 PM
    Merited by RetiredCoder (5)
     #19
    Code:
    import secp256k1 as ice
    pub = '03C8F139DAD58B4786D3649992849733C2A7626F011089E87508CCDF8B0758C493'
    lmda = 0x5363ad4cc05c30e0a5261c028812645a122e22ea20816678df02967c1b23bd72
    N = ice.N

    G = ice.scalar_multiplication(1)
    Q = ice.pub2upub(pub)   # pvk
    QE = ice.pub_endo1(Q)   # pvk * lmda % N
    QG = ice.point_subtraction( Q, G )  # pvk -1
    if QG == QE:    # pvk -1 = (pvk * lmda % N)
        # pvk -1 = (pvk * lmda % N)
        # pvk - pvk * lmda = 1 %N
        # pvk (1 - lmda) = 1  %N
        # pvk = 1 (1 - lmda)  %N
        print(f'[+] PrivateKey: {hex(pow((1-lmda), N-2, N))}')
    Code:
    [+] PrivateKey: 0xc6768f199574104ae1b75eab82b0cc1d2d2e9ee7d50637a574e27131e9301552

    I missed it indeed. Sad

  • RetiredCoder (OP)
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    January 15, 2025, 12:21:28 PM
    Merited by iceland2k14 (2)
     #20
    Quote from: iceland2k14 on January 15, 2025, 12:09:39 PM
    Code:
    [+] PrivateKey: 0xc6768f199574104ae1b75eab82b0cc1d2d2e9ee7d50637a574e27131e9301552
    I missed it indeed. Sad

    Correct!
    You are already the winner of another mini-puzzle so don't be sad Smiley
    I've solved #120, #125, #130. How: https://github.com/RetiredC
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