Strassen’s factoring algorithm 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 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?