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Would a factor besides the normal discrete logarithm problem increase or decrease the solving time?

$k = n \cdot g^a \mod P$

with given $k,g,P$ and the knowledge $P= 2 \cdot N \cdot f+1$, while $f$ can be a product out of other primes. The factor $n<P-1 \in \mathbb{N}$. The generator $g$ can generate a group with max size of $N$.
How can we solve this?

Harder than solving the normal: $k' = h^a \mod Q$, with h prime rooot of $Q$?


edit: forgot to mention at least two equations need to be solved with same $n$
$k' = n \cdot g^b \mod P$
or one without
$k' = k \cdot g^c \mod P$

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$k = n \cdot g^a \mod P$

How can we solve this?

It is trivial to find $(n, a)$ pairs that satisfy this relation; select an arbitrary $a$ and compute $n = k g^{-a} \bmod P$; that's a solution.

Now, that'll give you $ord(g)$ distinct solutions; if you have a specific solution in mind, you have no way to telling which one it is. However, depending on how the solution is used, it might not matter...

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  • $\begingroup$ Thanks for response, always forgetting your can just apply simple transformation at mod functions. I think in use case it does matter. There a 2nd equation is given with $k'=n \cdot g^b \mod P$. So $a,b$ with same $n$ need to be found or $k'=k \cdot g^c$ $\endgroup$ – J. Doe May 6 at 16:51
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    $\begingroup$ If you extend the question to "solve a solution $a, b, n$ to the simultaneous equations $k = n \cdot g^a \bmod P$, $k' = n \cdot g^b \bmod P$", then there are still multiple solutions; however finding one solution is now equivalent to the DLog problem (equivalent in the sense that an oracle to solve one problem allows you to solve the other) $\endgroup$ – poncho May 6 at 18:39

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