[Strassen's factoring algorithm](http://web.maths.unsw.edu.au/~davidharvey/talks/factoring.pdf) shows that $\text{FACTORING} \in \text{DTIME}(N^{\frac{1}{4}+o(1)})$, but if I'm not mistaken in my analysis it also uses a similar amount of space. By making a trade-off I think it is possible to show $\text{FACTORING} \in \text{DTISP}(N^{k+o(1)}, N^{\frac{1}{2}-k+o(1)})$ for $\frac{1}{4} \leq k \leq \frac{1}{2}$.


On the other hand, trial division up to $\sqrt{N}$ demonstrates $\text{FACTORING} \in \text{DTISP}(N^{\frac{1}{2}+o(1)}, \log(N)^{O(1)})$. The only extra space we need is to keep a counter and perform the divisibility test.


Is there any deterministic factoring algorithm known to be in $\text{DTISP}(N^{k + o(1)}, N^{o(1)})$ for $k \lt \frac{1}{2}$?


I know that [there is an $N^{\frac{1}{3}+o(1)}$-time algorithm due to Michael Rubinstein](https://rjlipton.wordpress.com/2014/10/18/a-new-provable-factoring-algorithm/) but I can't tell what the space usage would be. This would qualify as an example if the space can be made subexponential.


Otherwise, is trial division the best we can do in $N^{o(1)}$ space?


We won't be able to prove $\text{FACTORING} \notin \text{DTISP}(\log(N)^{O(1)}, O(\log(N))) = \text{L}$, since that would imply $\text{NP} \neq \text{L}$, an open problem.



Fastest deterministic factoring algorithm in subexponential space?
[Lehman's algorithm](http://www.ams.org/journals/mcom/1974-28-126/S0025-5718-1974-0340163-2/S0025-5718-1974-0340163-2.pdf) uses $O(N^{\frac{1}{3}})$ time $O(\log N)$ space. The algoritm is the following.


0) Check that $n$ is odd and $n > 8$.


1) Check that every $a = 2, \ldots, [n^{\frac{1}{3}}]$ is not a divisor of $n$.


2) For every $k=2, \ldots, [n^{\frac{1}{3}}]$ and for every $d= 0,1 \ldots, [n^{\frac{1}{6}} / (4 \sqrt{k})]$ check is the number
$$([\sqrt{4kn}]+d)^2- 4kn $$ a square of an integer. If this is the case then consider $A:=[\sqrt{4kn}]+d$ and $B:= \sqrt{A^2 - 4kn}$. Note that 
$$A^2 \equiv B^2 (\text{mod } n ).$$ 
Check: $1 < \text{gcd}(A \pm B, N) < N$.


If we have this inequality then of course we can factorize $N$. Otherwise $N$ is prime by a theorem in the paper. 


