RSA Duplicate-Signature Attack
You forgot step 4:


  
  $m$ and $s$ should each be primitive roots mod $p$ and mod $q$
  


$m$ is not a primitive root mod $p$.  We can easily deduce that (without the bother of actually performing any nontrivial math) from the relation you gave:

$$s^{e_p} = m \bmod p$$

for $e_p$ even; hence $m$ must be a quadratic residue, and hence cannot be primitive.

As $e_p$ is even and $e_q$ is odd, there is no solution to the CRT...

For your reference, $m$ will be a primitive root of a prime $p$ if, for every prime factor $r$ of $p-1$, we have $m^{(p-1)/r} \not\equiv 1 \bmod p$