Bitcoin puzzle transaction ~32 BTC prize to who solves it

[mcdouglasx]

I'm bored, nothing interesting is happening around these woods no more.  Nobody has new and exciting ideas no more.

Something on mind for some time now, I wonder why 160 bit? Why not keep adding coins to 165, 170 etc?

And is it really feasible brute forcing any key beyond 80 bit? With no pub exposed of course.

I guess from 81 to 159 will not be solved in the next decade, why bother keeping them in there when we can use them as extra incentive for higher ranges, like adding them to 165, 170, 175 etc?

Good luck to all.😴

It's not that we don't have new ideas, it's that we don't share them for our own interest, I thought about sharing how I reduce the bits from 130 to 107, in a range of 1000 possible pubkey, but I haven't done it, because the The time when you did something important and people contributed to the developer.
I don't have how to update my PC, so bit 100-107 is still too much for me. An example of what I am saying is keyhunt, it is a tool used by thousands or millions and has only received a little more than 10k in donations.
If I get to unlock a puzzle, I'll share it, while not because nowadays people don't even post the pk of the last puzzles solved.

[digaran]

I'm bored, nothing interesting is happening around these woods no more.  Nobody has new and exciting ideas no more.

Something on mind for some time now, I wonder why 160 bit? Why not keep adding coins to 165, 170 etc?

And is it really feasible brute forcing any key beyond 80 bit? With no pub exposed of course.

I guess from 81 to 159 will not be solved in the next decade, why bother keeping them in there when we can use them as extra incentive for higher ranges, like adding them to 165, 170, 175 etc?

Good luck to all.😴

It's not that we don't have new ideas, it's that we don't share them for our own interest, I thought about sharing how I reduce the bits from 130 to 107, in a range of 1000 possible pubkey, but I haven't done it, because the The time when you did something important and people contributed to the developer.
I don't have how to update my PC, so bit 100-107 is still too much for me. An example of what I am saying is keyhunt, it is a tool used by thousands or millions and has only received a little more than 10k in donations.
If I get to unlock a puzzle, I'll share it, while not because nowadays people don't even post the pk of the last puzzles solved.

If you can solve a puzzle with your idea, delete all social media accounts, remove yourself from online world, use a new pc with tor+vpn enabled to transfer the funds and never ever come back to this forum to post or share anything, and only solve the next puzzle after 4, 6 month.

If you like to live long. This is my advice to anyone.
However if you first share your idea and then puzzles start getting solved, nobody would look at your direction but if your idea could be used to crack any 256 bit key, then cryptocurrencies no more!

[kalos15btc]

I'm bored, nothing interesting is happening around these woods no more.  Nobody has new and exciting ideas no more.

Something on mind for some time now, I wonder why 160 bit? Why not keep adding coins to 165, 170 etc?

And is it really feasible brute forcing any key beyond 80 bit? With no pub exposed of course.

I guess from 81 to 159 will not be solved in the next decade, why bother keeping them in there when we can use them as extra incentive for higher ranges, like adding them to 165, 170, 175 etc?

Good luck to all.😴

i already find two puzzle with my methode of hash160 lowering the range of searching,and still im going to 130 135 my methode is working its just lowering and find the range in subed pub key to lower bit range 2 times yes when you find same 14 digits of hash160 than means that adress is in that range, but the search is not for one adress, i get a lot of addresses im trying , and no im not going to share the pk of any address i work hard for this, i advise you to try on 120 my methode, and get your bitcoin cash prize, ps: ill leave 50 satoshi on 130 for you, thank you

[digaran]

i already find two puzzle with my methode of hash160 lowering the range of searching,and still im going to 130 135 my methode is working its just lowering and find the range in subed pub key to lower bit range 2 times yes when you find same 14 digits of hash160 than means that adress is in that range, but the search is not for one adress, i get a lot of addresses im trying , and no im not going to share the pk of any address i work hard for this, i advise you to try on 120 my methode, and get your bitcoin cash prize, ps: ill leave 50 satoshi on 130 for you, thank you

So you already solved 120 and 125? Can we see their private keys? Since you have worked hard for them, and don't want to share them, you could sign a message with them without revealing them.

I advise you to grind all the puzzles with no mercy, don't even leave a single sat for anyone.
Kalos the merciless.😅

Ps, keep your pk ( whatever that means ) for your collection.


Can someone help to make this script working?

Code:

class EllipticCurve:
    def __init__(self, p, a, b, g_x, g_y, n):
        self.p = p
        self.a = a
        self.b = b
        self.g_x = g_x
        self.g_y = g_y
        self.n = n

    def add_points(self, p1, p2):
        # Point addition logic here
        pass

    def subtract_points(self, p1, p2):
        # Point subtraction logic here
        pass

    def scalar_multiply(self, point, scalar):
        # Scalar multiplication logic here
        pass

# Replace these values with your desired parameters
N = ...
P = ...
G = (..., ...)  # (x-coordinate, y-coordinate)

# Create an instance of the EllipticCurve class with your parameters
curve = EllipticCurve(P, 0, 7, G[0], G[1], N)

# Replace with your target public key and desired number of subtractions
target_public_key = "..."
num = 100

# Example of point subtraction loop
subtract_point = G  # Initialize subtract_point with base point G

for t in range(num + 1):
    # Perform point subtraction
    result = curve.subtract_points(subtract_point, G)

    # Convert the result to hexadecimal representation
    h = (result[0], result[1])

    # Print or store the result as needed
    print("Subtraction result for iteration {}: {}".format(t, h))

    # Update subtract_point for the next iteration
    subtract_point = result

# Note: You need to implement the add_points and subtract_points methods
# with the actual point addition and subtraction logic for your curve.

[lordfrs]

Code:

from sympy import mod_inverse
import secp256k1 as ice
pub=ice.pub2upub('Here Compressed Public Key')
N=0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141

k=mod_inverse(2,N)
neg1=ice.point_negation(ice.scalar_multiplication(1))


def ters(Qx,Scalar):
     ScalarBin = str(bin(Scalar))[2:]
     le=len(ScalarBin)
     for i in range (1,le+1):
        if ScalarBin[le-i] == "0":
            Qx=ice.point_multiplication(k,Qx)
        else:
            Qx=ice.point_addition(Qx,neg1)
            Qx=ice.point_multiplication(k,Qx)
     return ice.point_to_cpub(Qx)


for x in range(1,65536):
         f= (ters(pub,x))
         data= open(“halfpub.txt”,”a”)
         data.write(f+”\n”)
         data.close()

Note this is where you decide how many bits should be reduced  
for x in range(1,65536):
For example reducing 26 bits requires 67108864 to be generated, 1 of them is the correct 26 bit reduced key.

Will you release for public to use?

Yes this is my code, I can write bit reduction code in another way if you want. If you want, you can reduce bits by subtraction.


I'm developing a new algorithm for downgrading from a known bit range to the bit range I want. When I complete it by giving only 1 correct pubkey, I will share it here.

[lordfrs]

i already find two puzzle with my methode of hash160 lowering the range of searching,and still im going to 130 135 my methode is working its just lowering and find the range in subed pub key to lower bit range 2 times yes when you find same 14 digits of hash160 than means that adress is in that range, but the search is not for one adress, i get a lot of addresses im trying , and no im not going to share the pk of any address i work hard for this, i advise you to try on 120 my methode, and get your bitcoin cash prize, ps: ill leave 50 satoshi on 130 for you, thank you

So you already solved 120 and 125? Can we see their private keys? Since you have worked hard for them, and don't want to share them, you could sign a message with them without revealing them.

I advise you to grind all the puzzles with no mercy, don't even leave a single sat for anyone.
Kalos the merciless.😅

Ps, keep your pk ( whatever that means ) for your collection.


Can someone help to make this script working?

Code:

class EllipticCurve:
    def __init__(self, p, a, b, g_x, g_y, n):
        self.p = p
        self.a = a
        self.b = b
        self.g_x = g_x
        self.g_y = g_y
        self.n = n

    def add_points(self, p1, p2):
        # Point addition logic here
        pass

    def subtract_points(self, p1, p2):
        # Point subtraction logic here
        pass

    def scalar_multiply(self, point, scalar):
        # Scalar multiplication logic here
        pass

# Replace these values with your desired parameters
N = ...
P = ...
G = (..., ...)  # (x-coordinate, y-coordinate)

# Create an instance of the EllipticCurve class with your parameters
curve = EllipticCurve(P, 0, 7, G[0], G[1], N)

# Replace with your target public key and desired number of subtractions
target_public_key = "..."
num = 100

# Example of point subtraction loop
subtract_point = G  # Initialize subtract_point with base point G

for t in range(num + 1):
    # Perform point subtraction
    result = curve.subtract_points(subtract_point, G)

    # Convert the result to hexadecimal representation
    h = (result[0], result[1])

    # Print or store the result as needed
    print("Subtraction result for iteration {}: {}".format(t, h))

    # Update subtract_point for the next iteration
    subtract_point = result

# Note: You need to implement the add_points and subtract_points methods
# with the actual point addition and subtraction logic for your curve.

Code:

Pcurve = 2**256 - 2**32 - 2**9 - 2**8 - 2**7 - 2**6 - 2**4 -1 # The proven prime
N=0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141 # Number of points in the field
Acurve = 0; Bcurve = 7 # This defines the curve. y^2 = x^3 + Acurve * x + Bcurve
Gx = 55066263022277343669578718895168534326250603453777594175500187360389116729240
Gy = 32670510020758816978083085130507043184471273380659243275938904335757337482424

def ECsub(point1x,point1y,point2x,point2y):
 
    neg_yq = Pcurve - point2y
   
    return ECadd(point1x,point1y,point2x,neg_yq) # point1-point2

you can use ready ice library, it will be faster

Ice.pointnegatition finds the inverse of the point. It will be extracted with ice.pointadd

The simple logic of the subtraction is to subtract the cord y from the total point point

[james5000]

Is there a way to determine if a point on the curve is even or odd?

[albert0bsd]

Is there a way to determine if a point on the curve is even or odd?

if i have a satoshi each time that i see that question...

Sadly there is no way to determine if a point in the curve is odd or even, also there is no way to determine if bit in any position is 1 o 0 that would totally break ECDSA

I already try a lot of things to try to anwser that question, but none of them works and actually i really give up that way, my recomendation is not lose time in that topic (Unless you are full cryptographer with wide knowledge in that field)

Welcome to the club by the way.

[ing1996]

Is there a way to determine if a point on the curve is even or odd?

if i have a satoshi each time that i see that question...

Sadly there is no way to determine if a point in the curve is odd or even, also there is no way to determine if bit in any position is 1 o 0 that would totally break ECDSA

I already try a lot of things to try to anwser that question, but none of them works and actually i really give up that way, my recomendation is not lose time in that topic (Unless you are full cryptographer with wide knowledge in that field)

Welcome to the club by the way.

Hello 👋 all.
If we divide an odd number X by 2, as a result we will have a remainder of X.5, is it also impossible to determine whether there is a remainder at this point?.

[yoshimitsu777]

hello.For private key 2 the x and y values of pubkey are

c6047f9441ed7d6d3045406e95c07cd85c778e4b8cef3ca7abac09b95c709ee5
1ae168fea63dc339a3c58419466ceaeef7f632653266d0e1236431a950cfe52a

i change generator points of elliptic curve to the values above.
let us assume we want to bruteforce scan the puzzles 1 to 20.
end range of puzzle20 is 0x100000 and normaly we would scan 1:100000 but as i changed G I divide the range by 2
therefore new reduced scan range is 0x1 to 0x80000

they keys found are:
4
26
70
101
A30
1498
649B

now we multiply by 2 and get correct prvkeys for seven puzzles 4 7 8 10 13 14 16

we miss thirteen keys and i understand that it didnt find them because they are and our scan used even.

someone please show what are the exact steps to find also rest keys within that range?

please show example with private key 4 for understanding

[Denis_Hitov]

hello.For private key 2 the x and y values of pubkey are

c6047f9441ed7d6d3045406e95c07cd85c778e4b8cef3ca7abac09b95c709ee5
1ae168fea63dc339a3c58419466ceaeef7f632653266d0e1236431a950cfe52a

i change generator points of elliptic curve to the values above.
let us assume we want to bruteforce scan the puzzles 1 to 20.
end range of puzzle20 is 0x100000 and normaly we would scan 1:100000 but as i changed G I divide the range by 2
therefore new reduced scan range is 0x1 to 0x80000

they keys found are:
4
26
70
101
A30
1498
649B

now we multiply by 2 and get correct prvkeys for seven puzzles 4 7 8 10 13 14 16

we miss thirteen keys and i understand that it didnt find them because they are and our scan used even.

someone please show what are the exact steps to find also rest keys within that range?

please show example with private key 4 for understanding

Please post the code so it's more clear what you're talking about.

[citb0in]

a mathematical curiosity that maybe could help the puzzle:

all numbers even that respects this succession 4,10,16,22,28,34,40,46.....To infinity
divided by 3

plus the sum of +1 to the same number divided by 3, results in an integer, odd number.

Code:

target = 100
target_2 = 100+1 #= 101

t1= target//3  #= 33.333333333333336
t2= target_2//3  #= 33.666666666666664

r= t1+t2 # = 67

---snipp---

t1 = t2 = 33
r will not result in 67 as you said, r=66

[mcdouglasx]

a mathematical curiosity that maybe could help the puzzle:

all numbers even that respects this succession 4,10,16,22,28,34,40,46.....To infinity
divided by 3

plus the sum of +1 to the same number divided by 3, results in an integer, odd number.

Code:

target = 100
target_2 = 100+1 #= 101

t1= target//3  #= 33.333333333333336
t2= target_2//3  #= 33.666666666666664

r= t1+t2 # = 67

---snipp---

t1 = t2 = 33
r will not result in 67 as you said, r=66

33.333333333333336+33.666666666666664 = 67
use 1 "/" symbol, 2 " //" is for rounding, sorry.
even if you run the script it gives you like

pk decimal=67

03df9d70a6b9876ce544c98561f4be4f725442e6d2b737d9c91a8321724ce0963f

edit

and if you subtract 100-67= 33
you would get the division of 100/3 rounded to 33.

[Denis_Hitov]

a mathematical curiosity that maybe could help the puzzle:

all numbers even that respects this succession 4,10,16,22,28,34,40,46.....To infinity
divided by 3

plus the sum of +1 to the same number divided by 3, results in an integer, odd number.

Code:

target = 100
target_2 = 100+1 #= 101

t1= target//3  #= 33.333333333333336
t2= target_2//3  #= 33.666666666666664

r= t1+t2 # = 67

---snipp---

t1 = t2 = 33
r will not result in 67 as you said, r=66

33.333333333333336+33.666666666666664 = 67
use 1 "/" symbol, 2 " //" is for rounding, sorry.
even if you run the script it gives you like

pk decimal=67

03df9d70a6b9876ce544c98561f4be4f725442e6d2b737d9c91a8321724ce0963f

edit

and if you subtract 100-67= 33
you would get the division of 100/3 rounded to 33.

Unfortunately, this method does not work for all numbers:

target = 150
target_2 = 150+1 #= 151

t1= target/3 #= 50
t2= target_2/3 #= 50.333333333333336

r= t1+t2 # = 100.333333333333336
150 − 100.333333333333336 = 49.666666666666664

[james5000]

Unfortunately, if the division results in a decimal number, the result will be incorrect.

[citb0in]

33.333333333333336+33.666666666666664 = 67
use 1 "/" symbol, 2 " //" is for rounding, sorry.

in Python the // operator stands for floor division, that means it divides the first number by the second number and rounds the result down to the nearest integer.

[james5000]

33.333333333333336+33.666666666666664 = 67
use 1 "/" symbol, 2 " //" is for rounding, sorry.

in Python the // operator stands for floor division, that means it divides the first number by the second number and rounds the result down to the nearest integer.

It works for the private key, but what we have is the public key; in this case, Python doesn't round the result.

[mcdouglasx]

33.333333333333336+33.666666666666664 = 67
use 1 "/" symbol, 2 " //" is for rounding, sorry.

in Python the // operator stands for floor division, that means it divides the first number by the second number and rounds the result down to the nearest integer.

is just what I wanted to say.

33.333333333333336+33.666666666666664 = 67
use 1 "/" symbol, 2 " //" is for rounding, sorry.

in Python the // operator stands for floor division, that means it divides the first number by the second number and rounds the result down to the nearest integer.

It works for the private key, but what we have is the public key; in this case, Python doesn't round the result.

It's just what the script does (if the condition I wrote is met)

a mathematical curiosity that maybe could help the puzzle:

all numbers even that respects this succession 4,10,16,22,28,34,40,46.....To infinity
divided by 3

plus the sum of +1 to the same number divided by 3, results in an integer, odd number.

Code:

target = 100
target_2 = 100+1 #= 101

t1= target//3  #= 33.333333333333336
t2= target_2//3  #= 33.666666666666664

r= t1+t2 # = 67

---snipp---

t1 = t2 = 33
r will not result in 67 as you said, r=66

33.333333333333336+33.666666666666664 = 67
use 1 "/" symbol, 2 " //" is for rounding, sorry.
even if you run the script it gives you like

pk decimal=67

03df9d70a6b9876ce544c98561f4be4f725442e6d2b737d9c91a8321724ce0963f

edit

and if you subtract 100-67= 33
you would get the division of 100/3 rounded to 33.

Unfortunately, this method does not work for all numbers:

target = 150
target_2 = 150+1 #= 151

t1= target/3 #= 50
t2= target_2/3 #= 50.333333333333336

r= t1+t2 # = 100.333333333333336
150 − 100.333333333333336 = 49.666666666666664

Why do you want to apply to a number that is division of 3? 150/3= 50

[Denis_Hitov]

Why do you want to apply to a number that is division of 3? 150/3= 50

I took the number 150 just as an example.

For example Puzzle #65:

target = 30568377312064202855
target_2 = 30568377312064202855+1 #= 30568377312064202856

t1= target/3 #= 10189459104021400951.666666666666667
t2= target_2/3 #= 10189459104021400952

r= t1+t2 # = 20378918208042801903.666666666666667

30568377312064202855 − 20378918208042801903.666666666666667 = 10189459104021400951.333333333333333


I cannot understand how this method will help in solving the puzzle if the "target" is unknown to us.

[james5000]

Why do you want to apply to a number that is division of 3? 150/3= 50

I took the number 150 just as an example.

For example Puzzle #65:

target = 30568377312064202855
target_2 = 30568377312064202855+1 #= 30568377312064202856

t1= target/3 #= 10189459104021400951.666666666666667
t2= target_2/3 #= 10189459104021400952

r= t1+t2 # = 20378918208042801903.666666666666667

30568377312064202855 − 20378918208042801903.666666666666667 = 10189459104021400951.333333333333333


I cannot understand how this method will help in solving the puzzle if the "target" is unknown to us.

I'm also quite curious about it.