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In BLS multi-signature aggregation scheme we can combine different signatures of different peers into one single signature, and then we can verify whether the aggregated signature is truly signed by the claimed peers.

My questions is, having an aggregated signature, is it possible to separate the signatures and achieve the individual signatures of the peers? For example, having an aggregated signatures of A, B, and C, is it possible to separate it and get the original signature of A?

I would appreciate any resource.

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For example, having an aggregated signatures of A, B, and C, is it possible to separate it and get the original signature of A?

Well, in BLS, an aggregated signature just the sum of the original sigantures; $\text{sign}(A,B,C) = \text{sign}(A) + \text{sign}(B) + \text{sign}(C)$ (where $\text{sign}(A,B,C)$ stands for the aggregate signature of $A, B, C$, and the addition is done on the pairing-friendly elliptic curve)

Hence, if you're just given the aggregated signature $\text{sign}(A,B,C)$, there isn't enough information to recover one of the original signatures.

However, if you do have the signatures of $B$ and $C$, you can subtract them out to reconstruct the original signature of $A$; $\text{sign}(A) = \text{sign}(A,B,C) - \text{sign}(B) - \text{sign}(C)$

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  • $\begingroup$ Thank you for your answer, that makes sense. You without having others signatures, it's impossible to compute individual signature. $\endgroup$
    – Arian B
    Commented Mar 23, 2022 at 15:52
  • $\begingroup$ another question, if we have a signature in BLS, sign(A), and we have a big prime number, P, can we derive sign(A) having sign(A) ^ P? $\endgroup$
    – Arian B
    Commented Mar 23, 2022 at 15:54
  • $\begingroup$ @ArianB: yes, we can compute $P^{-1} \bmod q$ (where $q$ is the group order), and so $\text{sign}(A) = (\text{sign}(A)^P)^{P^{-1}}$ $\endgroup$
    – poncho
    Commented Mar 23, 2022 at 16:12

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