Simultaneous equations with discrete log problem I know discrete log problem is a hard problem with one, lets say  $m^{23} \mod 320 \equiv 300$
But would it be easy and possible to solve, lets say $m^{23} \mod 320 \equiv 300$
 and $m^{31} \mod 320 \equiv 261$ 
How does one go about solving it? If it is not possible, why not?

This system has no solution.
Given, $m^{23} \equiv 300 (\mod 320)$ and $m^{31} \equiv 261 (\mod 320)$.
Now, $m^{31} \equiv 261 (\mod 320)$
$\implies m^{23+8} \equiv 261 (\mod 320)$
$\implies m^{23} \times m^8 \equiv 261 (\mod 320)$
$\implies 300m^{8} \equiv 261 (\mod 320)$
$\implies 300m^{8} -320k=261$, for some integer $k$.
But this is not possible. Because the LHS of the equation is divisible by $10$, but the RHS isn't.
