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?
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7$\begingroup$ 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. $\endgroup$– Noam D. ElkiesAug 29, 2012 at 3:23
1 Answer
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.)
http://gdz.sub.uni-goettingen.de/dms/load/img/?PPN=GDZPPN00218186X&IDDOC=253620
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
http://www.ams.org/journals/tran/2001-353-08/S0002-9947-01-02782-9/