Given positive integers $1< x,y < 2019$ that satisfy system of equations
$1992^{x} \ \ (mod \ 2019) = n$ and $n^{y} \ \ (mod \ 2019) = 1992$
Question :
(i) How many pairs of $(x,y)$ satisfy above condition?
(ii) Let $S=\left \{ x \right \}$ and $T = \left \{ y \right \}$ What is the sum of members in $S$ and $T$
I created this Question from reading an RSA encryption ,but fundamentally I solved this and got only $(5,269)$ for one solution ,there are so many pairs of $(x,y)$ that take time to find out .
Are there any quicker way solution to manage$(x,y)$ and find the sum of members in set $S$ and set $T$?
I appreciate for any helps, Thank you.
You are looking for solutions to $(1992^x)^y=1992^{xy}\equiv 1992 \pmod {2019}$
By Euler's theorem we have $a^{\varphi(2019)}\equiv 1 \pmod {2019}$ for any $a$ coprime to $2019$. $\varphi(2019)=1344$. The order of $1992 \mod 2019$ must be a factor of $1344$ and we can easily check that $1992^{672},$ $1992^{348},$and $1992^{192}$ are not $1$, so you need $xy\equiv 1 \pmod {1344}$. Any $x$ coprime to $1344$ will have an inverse, which will be the corresponding $y$. For example, we have $13^{-1}\equiv 517 \pmod {1344}$ and $1992^{13 \cdot 517}=1992^{6721}\equiv 1992 \pmod {2019}$ so $(13,517)$ is another pair.