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Jul 9, 2018 at 17:41 comment added irakliy I've actually created a separate question for this here.
Jul 9, 2018 at 16:51 comment added irakliy Thank you - this makes sense. If I can force the value $x$ to be always smaller than the modulus of the "smaller" ring - this shouldn't be an issue, right? I'm thinking this could be done with the methodology provided in the answer if I require that $\frac{v}{u} < q$ (this would imply that $x < q$).
Jul 9, 2018 at 16:28 comment added Vadym Fedyukovych A "smaller" ring element can be considered "equivalent" to more than one element of the "larger" ring, defined by the larger modulus. It is reasonable to ask for "equality of logarithms" in groups of the same order, or in the case of a group of a hidden order.
Jul 8, 2018 at 17:27 comment added irakliy What does ill-defined mean in this context?
Jul 8, 2018 at 10:08 comment added Vadym Fedyukovych This equality would be ill-defined in case of two different known orders of groups involved.
Jun 19, 2018 at 17:48 vote accept irakliy
Jun 19, 2018 at 7:15 answer added Changyu Dong timeline score: 4
Jun 19, 2018 at 5:47 history asked irakliy CC BY-SA 4.0