How fast would a nondeterministic algorithm for an EXPTIME-complete problem have to be to imply $P \neq NP$? How fast would a nondeterministic algorithm for an EXPTIME-complete problem have to be to imply $P \neq NP$?  A polynomial time nondeterministic algorithm would immediately imply this because $P \neq EXPTIME$ but no one believes $NP = EXPTIME$.  If I've done the algebra right (see below) the time hierarchy theorem would still give the $P \neq NP$ implication for $O(2^n/f(n))$ running times for any superpolynomial $f(\cdot)$, but for all I know there are complete problems with efficient reductions that allow slower algorithms to give the result.  Are there EXPTIME-complete problems where we know something like $2^n/n$ or $2^n/n^2$ with nondeterminism is enough?

Clarification of the "algebra": $P = NP$ implies $EXPTIME=NEXPTIME$ by a padding argument, so a nondeterministic $2^n/f(n)$ algorithm for an EXPTIME-complete problem would also be one for a NEXPTIME-complete problem.  For superpolynomial $f(\cdot)$ this would contradict the nondeterministic time hierarchy theorem since we could reduce using some $L \in$ NTIME$(2^n)$.
I think its easier to turn it around.

If $\mathsf{P}=\mathsf{NP}$, then $\mathsf{NTIME}(T(n)) \subset \mathsf{DTIME} ((T(n))^c)$ 
for some constant $c$, and any $T(n) &gt; n$.
Since $\mathsf{DTIME} ((T(n)^c)$ does not contain 
$\mathsf{DTIME}(T(n)^{c}\log T(n)) \subset \mathsf{DTIME}(T(n)^{c+1})$, 
this means we cannot solve, say all problems in $\mathsf{DTIME}(2^n)$ in 
$\mathsf{NTIME} (2^{\epsilon n})$ for some $\epsilon$.
So a non-deterministic time $2^{o(n)}$ algorithm for a problem complete for
$\mathsf{DTIME} (2^n)$ under quasi-linear reductions would be enough to prove $\mathsf{P} \neq \mathsf{NP}$.