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I'm working on a problem that gives me two private keys $d_1,d_2$ that work with the same $(N,e)$ pair. The problem gives me the two private keys and $N$. Is it possible to find $p$ and $q$? Or how would I find $p$ and $q$ or the sum $p+q$?

SEJPM
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  • A single private key and the matching public key is enough to factor the public modulus $N$; see this. This is why it is a bad idea that several public keys share the same modulus $N$. Further, if several private keys match the same public key $(N,e)$, then these private keys are functionally equivalent (they need not be equal). – fgrieu May 22 '18 at 13:23
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    But the public key is not given –  May 22 '18 at 13:57
  • The question as stated makes no sense: for fixed $N$ and $e$ there is only one possible value of the private key $p$, $q$, and $d$ (up to equivalence modulo $\lambda(N) = \operatorname{lcm}(p - 1, q - 1)$). – Squeamish Ossifrage May 22 '18 at 14:59
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    @SqueamishOssifrage: "up to equivalence"; I suspect that's the trick that textbook is looking for... – poncho May 22 '18 at 15:11
  • But this problem has already been solved before, I'm just wondering how it was done. –  May 22 '18 at 17:48
  • I forget where the question was, but someone said that for RSA you only need a multiple of (p-1)(q-1) –  May 22 '18 at 17:51
  • Write down the relation between $d$ and $e$, and the problem practically solves itself… – Squeamish Ossifrage May 22 '18 at 20:30
  • I solved it. d2 - d1 = n+1-(p+q) –  May 23 '18 at 00:31

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