In RSA: Fast factorization of N if d and e are known a comment under the OPs question stated that if the encrytion exponent $e$ is small compared to $N=p\cdot q$ for the RSA-primes $p,q$ (like $e<\sqrt{N}$) one could find $k$ in $de=1+k\varphi(N)$ by rounding $\frac{de}{N}$ as $k$ is an integer and $\varphi(N)\approx N$.
I don't know how to show this rigorously as I don't find a way to use $e<\sqrt{N}$ and I'm unsure whether the floor function or some other approximation was meant. (For the following i went with the floor as it seemed to be the most reasonable)
As $\varphi(N)\le N$ I know that: $$k=\left\lfloor \frac{1}{N}+k\right\rfloor\ge\left\lfloor \frac{de}{N}\right\rfloor$$ As $\varphi(N)=N-p-q+1$ is also know that: \begin{align*} \left\lfloor\frac{de}{N}\right\rfloor&=\left\lfloor\frac{1}{N}+k\frac{\varphi(N)}{N}\right\rfloor\\ &=\left\lfloor\frac{1}{N}+k\frac{pq-p-q+1}{pq}\right\rfloor\\ &=\left\lfloor\frac{1}{N}+k\left(\frac{1}{N}-\frac{1}{q}-\frac{1}{p}\right)\right\rfloor+k \end{align*} So it would be nice to have $\left\lfloor\frac{1}{N}+k\left(\frac{1}{N}-\frac{1}{q}-\frac{1}{p}\right)\right\rfloor=0$ as this would show the preposition.
Does this result even hold true?
A small $e$ is not an issue for factoring $N$, a relatively small $d$ is (this can be attacked by the continued fraction technique, or lattices). Typically in modern systems $e=2^{16}+1$ is quite common, as it's prime and has only two bits set, so makes for relatively fast encryption or signature checking. If both $e$ and $d$ are known, there is a fast probabilistic attack to factor $N$ (as you linked to). So $d$ really has to be kept secret (of course) and cannot be chosen conventiently small (for fast decryption, say); this would typically make $e$ large (and suspicious: $e$'s that are not small stand out, and are inefficient ) and attackers would probably break $d$ and thus $N$ and the system.. A small $e$ alone gets you no advantage...