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Show that if two different RSA public keys $p_k$s are known to an attacker for the same secret key $s_k$, then $s_k$ can be broken

I've deduced that that if the 2 public key exponents are $e_1,e_2$ then they have the same remainder modulo $\phi$, but that still doesn't help me determine $d$.

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  • $\begingroup$ Hint: the common secret key $S_k$ is $(n,d)$. A public key $p_k$ is $(n,e)$. That's the same $n$. $\endgroup$
    – fgrieu
    Commented Oct 3, 2022 at 8:22

2 Answers 2

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This is a homework question, and so I'll give you a hint, not the answer.

The intended approach to take is not to recover $d$ directly; instead, it is to factor the modulus $n$ (and once you have that, recovering $d$ is easy).

So, if you have the value $n$ and the value $k \phi(n)$ for some unknown integer $k$, how could you factor?

One simple approach would work if you assume $k$ isn't too large. There are fancier approaches where you don't have to make that assumption, but why don't you start with the simplifying assumption...

(BTW: you actually have the value $k \lambda(n)$ for $\lambda(n) = \text{lcm}(p-1, q-1)$, however that doesn't really matter for this question...)

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You should take a look at how $p_{k_1}$ and $p_{k_2}$ are derived from $s_k$ (or more concretely how $e_1$, $e_2$ are derived from $d$). Once You understand it, see if there is a way to deduce the modulus of this operation and from this compute the $s_k$.

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