N-Engel and torsion-free threshold at least n implies solvable iff nilpotent for Lie rings

Statement
Suppose $$L$$ is a Lie ring and $$n$$ is a natural number such that:


 * 1) $$L$$ is a $$n$$-Engel Lie ring, i.e., for any $$x \in L$$, we have $$(\operatorname{ad} x)^n = 0$$ where $$\operatorname{ad}$$ denotes the adjoint action.
 * 2) The torsion-free threshold]] of $$L$$ is at least $$n$$, i.e., $$L$$ is $$p$$-torsion-free for all primes $$p \le n$$.

Then, $$L$$ is a solvable Lie ring if and only if it is a nilpotent Lie ring. Note that nilpotent implies solvable for Lie rings, so the informative direction is the solvable implies nilpotent direction. Explicitly, if $$L$$ has derived length $$r$$, then it is nilpotent and its nilpotency class is at most:

$$\frac{n^r - 1}{n - 1}$$

Note, conversely, that from general information, if $$L$$ has nilpotency class $$c$$, the derived length is at most $$\log_2(c + 1)$$.

Related facts

 * 2-Engel implies class three for Lie rings
 * 2-Engel and 3-torsion-free implies class two for Lie rings
 * 3-Engel and (2,5)-torsion-free implies class six for Lie rings
 * 4-Engel and (2,3,5)-torsion-free implies nilpotent for Lie rings