# Erdős-Straus with 4 terms

The Erdős-Straus conjecture states that any fraction of the form $\frac{4}{n}$ can be decomposed as an Egyptian fraction with just 3 terms. In related research, I've recently come across conditions on $n$ (e.g. if $n$ has factorization of such and such a form) so that $\frac{k}{n}$ can be decomposed as an Egyptian fraction with $k$ terms. This isn't strong enough to prove Erdős-Straus, but I was wondering if there were already results like this out there or if this is new. It seems like most of what I read examines the 3-term decompositions. Are 4-term decompositions trivial to come by?

-
It's known that any fraction $k/n$ is the sum of at most $k$ terms $1/m_i$ by the "greedy algorithm": just let $m_1 = \lceil n/k \rceil$ (i.e. so $1/m_1$ is maximal given $1/m_1 \leq k/n$), check that $k/n - 1/m_1$ has numerator at most $k-1$, and proceed by induction. –  Noam D. Elkies Aug 29 '12 at 3:23

Sums of 4 (or generally $k$) unit fractions are by no means trivial.

There is a general criterion to Y. Rav (On the representation of rational numbers as a fixed sum of unit fractions. J. Reine Angew. Math. 222 (1966): 207-213.)

The equation $\frac{m}{n}= \frac{1}{x_1} + \cdots + \frac{1}{x_k}$
is certainly soluble for $m \leq k$, but for $m>k$ the same type of problems arise that one has for the Erdos-Straus equation.
One can expect that for fixed $m$ and fixed $3 \leq k < m$, there is some finite bound $N_{m,k}$ such that $n>N_{m,k}$ admits a solution. But this is an open problem.
One can prove that for "almost all" $n \leq N$. The strongest version of "almost all", and a discussion of the parametrization of such equations is in C. Elsholtz, Sums of $k$ unit fractions, Trans. Amer. Math. Soc. 353 (2001), 3209-3227