Erdos-Straus with 4 terms

2012-08-29T02:43:07Z

The Erdos-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 Erdos-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?

Answer by Christian Elsholtz for Erdos-Straus with 4 terms

2012-08-29T07:46:43Z

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/

Erdos-Straus with 4 terms - MathOverflow

Erdos-Straus with 4 terms

Answer by Christian Elsholtz for Erdos-Straus with 4 terms