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?
 A: It is unsolvable: $\ {\rm mod}\ 2\!:\ x^n\equiv x\,\Rightarrow\, m\equiv m^{23}\equiv300\equiv\color{#c00}  0\ $ contra $\ m\equiv m^{31}\equiv 261\equiv\color{#c00} 1 $

If instead the modulus is coprime to $\color{#0a0}{300}$ or $\color{#c00}{261}$ you could then argue either
$\,\qquad 31(3)-23(4)\, = \,1\ $  so $\ m \equiv  (m^{\large 31})^{\large 3} \ \, (m^{\large  23})^{\large  -4}\, \equiv\, 261^{\large 3}\  \color{#0a0}{300^{\large  -4}}\equiv\ \ldots$
or $\,\ 23(27)-31(20) = 1\ $ so $\ m \equiv  (m^{\large 23})^{\large 27} (m^{\large  31})^{\large  -20} \equiv 300^{\large 27} \color{#c00}{261^{\large  -20}}\equiv\ \ldots$
and then verify if the computed value is indeed a solution.
A: 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.
