# How to derive the Private key when the two the nonces are k and k*2

Is the same private key using k and k*2 respectively in two different signatures vulnerable? If yes How to calculate the key in this case

Here is an example, Both signatures are valid and the nonce of r2,s2,z2 is double the first signature's nonce, how do I calculate private key from these two signatures?

r1 = 0xdcf17de661e280dbf62e03ef1655d1baaabc301da9fc6b29a63e52e7780c115d

s1 = 0x20be91ddd5598e22fa43014172df5312275fbdb462a2e9855c7a7433138a4a9c

z1 = 0xbb1e00d2027efd3085b83de2a3602a8ea49e0c9d5b821cd6291d5feefd410303

s2 = 0x4f08689cd10e8fd5e9be108d18e2ee582774dc409176643800284669d6e787eb

z2 = 0x0fa25a9a0712366012de680bed93f8c94ffa2c119464d1e59b0a03b1c115b54f

• Are you using the RFC6979 standard for deriving k? Nov 27, 2021 at 20:31
• How to do that ? Nov 27, 2021 at 21:23
• Read the RFC: datatracker.ietf.org/doc/html/rfc6979 Nov 28, 2021 at 15:59
• I've rolled back the edit of the question post to return it to the question. Please refrain from disparaging other users in posts.
– Murch
Dec 3, 2021 at 4:59

c0e2d0a89a348de88fda08211c70d1d7e52ccef2eb9459911bf977d587784c6e is the k1 nonce. a3c2dce777813dc3f0d7a105a5fd56d5894ba30ffa01ba79667cadba37a9d8df is the private key.

• I don't see how you can compute this based on the information in the question. Can you elaborate on the technique you used? Dec 1, 2021 at 16:13
• I think someone put together this question or a challenge somewhere else to lead up to this kind of reveal. The "nonce" value `c0e2d0a89a348de88fda08211c70d1d7e52ccef2eb9459911bf977d587784c6e` appears in multiple blog posts about "hacking bitcoin wallets".
– Murch
Dec 1, 2021 at 19:31
• The private key posted here previously appeared in this stack exchange post before: bitcoin.stackexchange.com/q/37740/5406
– Murch
Dec 3, 2021 at 5:01

Assuming you're talking about ECDSA signing. For BIP340/Schnorr signatures, see this answer.

### For valid signatures

Let:

• `d`: the private key
• `P=d*G`: the public key
• `k1`: the (first) nonce
• `R1=k1*G`: the public first nonce
• `r1=R1.x mod n`: the public first nonce as it will be encoded in the signature.
• `k2=2*k1`, `R2=k2*G=2*R1`, `r2=R2.x mod n`: the same for the second nonce
• `z1` and `z2`: the respective message hash

The two signatures will then be the pairs `(r,s)` and `(r',s')` for which:

• `s1 = (z1 + r1*d) / k1 mod n`
• `s2 = (z2 + r2*d) / k2 mod n = (m2 + r2*d) / (2*k1) mod n`

Multiplying both sides of the equations by their denominator on the right hand side:

• `s1*k1 = z1 + r1*d mod n`
• `2*s2*k1 = z2 + r2*d mod n`

Assuming `2*r1*s2 ≠ r2*s1`, this is a set of two linear equations in two unknowns (`k1` and `d`), with solution:

• `d = (z2*s1 - 2*z1*s2) / (2*r1*s2 - r2*s1) mod n`
• `k1 = (z2*r1 - z1*r2) / (2*r1*s2 - r2*s1) mod n`

Don't ever use related nonces for ECDSA (or Schnorr) signatures. Create a fresh, independent, nonce every time. The industry standard is to generate nonces using RFC6979.

### For fake "signatures"

The values you've provided however do not correspond to real signatures, despite satisfying the equation. That's because for a signature to be valid, you have to give the messages which hash to the `z` values, not just the result. And you can't do that here, because `z2 = (r2/r1)*z1`. Such matching ratios will not (and cannot) occur for `z` values that are the result of hashing.

It turns out that given any ECDSA triplet `(r,s,z)`, and integer `a`, another triplet can be found. Let `R` be an elliptic curve point whose X coordinate equals `r` (mod n), and `R' = aR`, and `r'` the X coordinate of `R'`. Then `(r',r'*s/(a*r'),r'*z/r)` is another ECDSA triplet, corresponding to multiplying `k` with `a`. However, again, this is not a valid signature because its z value will not be the hash of a computable message. And the formula above will also fail to retrieve the private key in this case, because the second "signature" wasn't created by using the private key.

• Thanks for your answer, I edited the question and provided the example , Have a look and if you answer according to that it will be easy for me to understand Nov 29, 2021 at 7:07
• I've updated my answer. Your input values don't correspond to actual signatures, and you cannot recover the private key from them. Nov 30, 2021 at 17:19
• pk = ((k1 - modinv(s2, n)*k2 + modinv(s1,n)*k1) * (modinv(modinv(s2,n)*r2 - modinv(s1,n)*r1, n)) % n) ,Is it possible to rewrite it based on the k and k*2 rule Dec 1, 2021 at 5:56
• Sure, just substitute k2=2*k1. But you'll find that for the specific numbers you've given above, the formula you get doesn't let you compute the private key (you get a division by zero). Dec 1, 2021 at 16:24
• but `z` can be just `2` or so. like in eth. say id of the method. then this becomes possible Dec 3, 2021 at 8:53