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We are given a number of distinct RSA public keys $(e_i,N_i)$ sharing the same unknown decryption exponent $d$. The $e_i$ were computed from $d$ drawn first, so we have different public keys. Can we recover $d$ in any manner?

(Edit) We also know $d$ is 333-bit. $N$ is the product of $p$ and $q$ which both are 512-bit prime numbers, making $N$ 1024-bit. The $e_i$ have been calculated from $d$ by making use of the equation: $e_i \cdot d \cong 1\mod \phi(N_i)$. So we have 3 sets of public keys $(N_i,e_i)$. I was asking if it possible, any way, to recover the value of $d$ if the maximum value of $i = 57$ (Means we can have at most 57 sets of public keys).

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    $\begingroup$ Welcome to crypto-SE! For 1024-bit $N_i$, a 333-bit $d$ is slightly above the 299-bit bound of Dan Boneh & Glenn Durfee's Cryptanalysis of RSA with Private Key $d$ Less than $N^{0.292}$. therefore that or M. Wiener's Cryptanalysis of short RSA secret exponents would need an extension. [Update: I do not know such extension, nor immediately found one!] $\endgroup$
    – fgrieu
    Commented Feb 11 at 11:28
  • $\begingroup$ Can you show Wiener's attack extension for this set? $\endgroup$
    – Pyp
    Commented Feb 11 at 11:45
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    $\begingroup$ This can be solved with a generalization of the lattice version of Wiener's attack. There's an explicit writeup here. $\endgroup$ Commented Feb 12 at 2:59
  • $\begingroup$ @SamuelNeves thank you, the paper works and seems to be the idea I want. The issue is I don't know how to fully place the answer here. For someone who is good at reading papers, please assist the community. $\endgroup$
    – Pyp
    Commented Feb 12 at 4:41
  • $\begingroup$ @Pyp: You should be able to answer your own question by typing text in "Your Answer" below then clicking "Post Your Answer" at the bottom. You need to be logged-in. You'll even get rep points if the answer is upvoted. $\endgroup$
    – fgrieu
    Commented Feb 12 at 9:55

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