We study proportional cake cutting under an equal-length constraint, where each of $n$ agents must receive a bundle of length exactly $1/n$ that is worth at least $1/n$ to that agent. Our main result determines the exact worst-case number of cuts needed to guarantee such an allocation, namely $2n-2$ for every $n$. This improves on the previously known quadratic upper bound. The proof combines the constrained necklace-splitting theorem with a recursive assignment argument based on Hall's theorem. The same bound extends to the stronger requirement of envy-freeness when $n$ is a prime power, and an $O(n^{1.525})$ upper bound holds for arbitrary $n$. We then ask whether such allocations can be computed. We first show that no deterministic protocol in the Robertson--Webb model terminates after finitely many queries on every instance and always returns an exact equal-length proportional allocation, even for two agents with identical valuations. For every $\varepsilon>0$, however, $O(n^2/\varepsilon)$ queries suffice to produce an allocation in which every agent receives a bundle of length exactly $1/n$ and value at least $1/n-\varepsilon$, again using at most $2n-2$ cuts. Finally, we augment the Robertson--Webb model with two-knife queries. We show that recursive balanced halving attains the optimal bound of $2n-2$ cuts when $n$ is a power of two. For arbitrary $n$, we give a procedure using $O(n\log n)$ cuts.

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