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Given a generator $g$, a large, safe prime $P$ and a result of the DH key exchange $g^{xy} \mod P$, how would I come up with two different $x', y'$ s.t. $g^{x'y'} = g^{xy} \mod P$

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  • $\begingroup$ $x$ and $y$ are random. if $x$ or $y$ are factorable then you can have. $\endgroup$ – kelalaka Dec 6 '18 at 9:40
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    $\begingroup$ Totally trivial answer: $x' = xy$, $y'=1$ $\endgroup$ – poncho Dec 6 '18 at 15:07
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In DHKE, Alice and Bob chooses random $x$ and $y$, $1 \leq x,y < n$, respectively.

For simplicity assume that $x$ is a product of two primes $x = r \cdot s$

Then set $ x' = r \neq x$ and $y' = s \cdot y \neq x$ then $g^{x'y'} = g^{rsy}=g^{xy}$ will be another pair.

if $x$ and $y$ are primes then there are two cases $xy < \varphi(P)$ and $xy > \varphi(P)$ that will result in the same equation that need to be solved.

$$xy \equiv x' y' \mod \varphi{(P)}\;\;, 1 \leq x',y < n$$

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  • $\begingroup$ Is it possible if $x$ and $y$ are prime? $\endgroup$ – dc3cdd1fc7 Dec 6 '18 at 9:50
  • $\begingroup$ there can be due to the modulus operation. You can use this to find a case. $\endgroup$ – kelalaka Dec 6 '18 at 10:06

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