Nonce k and k +1 (ECDSA SIGNATURE)

[NotATether]

How can one calculate a message hash?

You take the text input, whether it is some signed message text or a raw transaction, and then pass it through SHA256(SHA256(input)), and save the result as e.

For ECDSA done on the secp256k1 curve, such as all kinds of bitcoin signatures, we are done at this point and we can set h = e our message hash.

However for some other different curves, you have to take the leftmost n bits of e after the double SHA256 hash. Where n can be found from the group order of the curve. So for example, secp256k1's group order is about 2^256 so it's n would be 256.

Keep in mind that the leftmost n bits equals the entire length of e if log2(curve's group order) == bit length of hash function used. When we are using both double SHA256 and secp256k1 curve in ECDSA like we are now, we know the double SHA256 always outputs 256 bits so these two values are equal and the entirety of e is used as the message hash.

The above paragraph implies that you are able to use a different hash function other than double SHA256 provided that its number of bits of output is greater than log2(curve's group order), because smaller-length hash functions are not allowed to be used with larger curve orders.

[1714749044]

1714749044

[ecdsa123]

explain what do you  mean " how u do this". explain what do this?what about talking? please be precisly when you ask about something. we are not in your head. we don't know what about you thinking when you write something.

[NotATether]

r1/s1 mod order = r2/s2 mod order

it's same signature, no diffrent signature

Code:

k1 == 109263722787838616791900575947640359553086907200677310074463510255775504782173*x + 33373073398809441106621025265904429856170478887328914010434069704980389675914
k2 == 109263722787838616791900575947640359553086907200677310074463510255775504782173*x + 33373073398809441106621025265904429856170478887328914010434069704980389675915

sage: r1/s1%N
109263722787838616791900575947640359553086907200677310074463510255775504782173

sage: r2/s2%N
109263722787838616791900575947640359553086907200677310074463510255775504782173

Can u give the sagemath or python code on how u do this.. Thank u so much.

I have never used sagemath before but this is sage code so it should work exactly as-is.

It might also work in Python just like that, considering that it uses GMP under the hood, but you might see exponents to the power of 10  instead of the actual number.

[Kpot87]

How can one calculate a message hash?

You take the text input, whether it is some signed message text or a raw transaction, and then pass it through SHA256(SHA256(input)), and save the result as e.

For ECDSA done on the secp256k1 curve, such as all kinds of bitcoin signatures, we are done at this point and we can set h = e our message hash.

However for some other different curves, you have to take the leftmost n bits of e after the double SHA256 hash. Where n can be found from the group order of the curve. So for example, secp256k1's group order is about 2^256 so it's n would be 256.

Keep in mind that the leftmost n bits equals the entire length of e if log2(curve's group order) == bit length of hash function used. When we are using both double SHA256 and secp256k1 curve in ECDSA like we are now, we know the double SHA256 always outputs 256 bits so these two values are equal and the entirety of e is used as the message hash.

The above paragraph implies that you are able to use a different hash function other than double SHA256 provided that its number of bits of output is greater than log2(curve's group order), because smaller-length hash functions are not allowed to be used with larger curve orders.

you want say that H(m) of transaction? if message is empty/null it just sha256(sha256(020000000155010ca6a15764977be218d19259d3e021b80851a1338530ad40d612f07c4b5801000 0006a47304402202062fb0a71961e18f155a3d54b468f1560425a8bd8a7fc9c6064aac149a24108 02201e5469c0d89bb32faf087eabf1d631c35b829bc45e09a8728e88c320811b01fc01210248d31 3b0398d4923cdca73b8cfa6532b91b96703902fc8b32fd438a3b7cd7f55ffffffff019808000000 0000001600143faaa7380c35d3d307b7caa3d2a1038fd3fe2c0500000000)) - and thats it?

[Bglhn]

Hello friends. I have two RSZ values obtained from the transfer of a bitcoin address and I want to find the nonce/private key. I do not write the values for confidentiality reasons, but I give approximate values as an example. k can be k+1. I'm sure there are many people here who can figure this out, but I can't. Any python code formulas etc that can help me? is there? I need your ideas.
R1 = 00a61d1110016763ed34995c319a42ea81b96a593efb29a4a46880bd8fe955077f
S1=009a72c80ae72e6edbe93d96d0202cc73bdf4ed1630c23381b2891e2427393878
Z1=306801f94f8bed2d753a66c60a614f359ff94758937bc7f950a9865d33ce1092

R2 = 00a6f4e7382a1c878a740e113c313779bcaa2dc20af5c1ff6c2bb7011cfb278c0d
S2=009cbddcba33bd30b4caad188ab02552e68b74fd43946e5b5a7f593dd367a26d28
Z2=9c76db1673ded5f0028abe36ad3b47bc47973681530481a32e1e7dd2f66ba0fd

The values are here, R1-R2 and S1-S2 are close to each other. I don't know how to make the connection. And of course how to calculate this correctly. I would be grateful if you help.

[Bglhn]

Hello friends. Inspired by the table on iceland's rsz, I created code in phyton for the cases where k,k+1...k+m. Since there is no such example whose private key I know, I ask you to check it. If it works, I will publish it on github. As far as I can see, there is no such resource, we can all benefit from it. Can anyone who sees it please check it out and give their opinions?

Code:

def h(n):
    return hex(n).replace("0x","")

def extended_gcd(aa, bb):
    lastremainder, remainder = abs(aa), abs(bb)
    x, lastx, y, lasty = 0, 1, 1, 0
    while remainder:
        lastremainder, (quotient, remainder) = remainder, divmod(lastremainder, remainder)
        x, lastx = lastx - quotient*x, x
        y, lasty = lasty - quotient*y, y
    return lastremainder, lastx * (-1 if aa < 0 else 1), lasty * (-1 if bb < 0 else 1)

def modinv(a, m):
    g, x, y = extended_gcd(a, m)
    if g != 1:
        raise ValueError
    return x % m
    
N = 0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141



R1 = 0x49bf1b1c8364c4179bd82a3be28b1a326c2c1b2d120c3264865ecbc4dbaed4b3
S1 = 0x4ad0d60d72880bf0a51d88d0d5138ffa3593273bd0b3d48a5afe04023db9c2c9
Z1 = 0x1362d682d8872a0451e5f0d86f743a62bf0730b57ddddc901668d837cbfa2f48
R2 = 0x00ad7991e3b3d36f6f17a22fad1faddc53e7c124e5b6626db172c79299fce5cfb6
S2 = 0x10ecd8352675027f74edc18180ac083d75a1488497c6c3078a5966015514ac46
Z2 = 0xfec02a5d53eb20a6e470b7c321e0da83ea6d677f600f67033abf4b0e6b8745aa
m = 1

print (h(((S2*m*R1 + Z1*R2 - Z2*R1) * (S1*R2 - S2*R1)^(-1)) % N))