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The solution to the problem of finding the primes $p$ and $q$ that have the following form:

$$ c_1 = (a_1 p + b_1 q)^{e_1} \mod{N} \\ c_2 = (a_2 p + b_2 q)^{e_2} \mod{N} $$

Given the values of $c_i$, $a_i$, $b_i$, $e_i$ and $N$ where $N = pq$.

Can be found in this writeup online. I follow the solution right up to the very last line where the author concludes that from the expression.

$$ a_2^{-e_1e_2}c_2^{e_1}-a_1^{-e_1e_2}c_1^{e_2}=a_2^{-e_1e_2}(b_2q)^{e_1e_2}+a_1^{-e_1e_2}(b_1q)^{e_1e_2} \mod{N} $$

We can find $q$ as:

$$ q = \gcd\left( a_2^{-e_1e_2}c_2^{e_1} - a_1^{-e_1e_2}c_1^{e_2} \mod{N}, N \right) $$

Perhaps because the expression has gotten so unweildy I've missed something simple but this seems like such a massive leap and there is no additional explanation.

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    $\begingroup$ $$a_2^{-e_1e_2}c_2^{e_1}-a_1^{-e_1e_2}c_1^{e_2}=a_2^{-e_1e_2}(b_2q)^{e_1e_2}+a_1^{-e_1e_2}(b_1q)^{e_1e_2} \mod{N}$$ $$a_2^{-e_1e_2}c_2^{e_1}-a_1^{-e_1e_2}c_1^{e_2}= q^{e_1e_2}\left( a_2^{-e_1e_2}(b_2)^{e_1e_2}+a_1^{-e_1e_2}(b_1)^{e_1e_2}\right) \mod{N}$$ $$a= q b \mod{N}$$ $$ab^{-1}= q \mod{N}$$ $\endgroup$
    – kelalaka
    Oct 15, 2023 at 19:12
  • $\begingroup$ @kelalaka Apologies for the topic necromancy, but I'm having a little bit of trouble seeing the relationship between your second line and your third. I see the format of $\textrm{expr} = \left(q^\textrm{exp}\right) \cdot \left(\textrm{expr}\right) \pmod{N}$, but I don't understand why the exponent on the $q$ can be dropped (or, honestly, how that results in a solution of $q$ being equal to a gcd). $\endgroup$
    – apnorton
    Apr 6 at 5:30
  • $\begingroup$ @apnorton It is just renaming. Once you know $c_1, c_2, a_1, a_2, e_1, e_2, N$ then you have the $q$ $\endgroup$
    – kelalaka
    Apr 6 at 19:20

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