This year's seventh-grade olympiad sponsored by Tel Aviv University, round two, held a couple of days ago, had this (translated by yours truly) as its third question:
Find four distinct natural numbers, $a$, $b$, $c$, and $d$, such that $$\frac1a+\frac1b+\frac1c+\frac1d=\frac37$$
My solution:$$\begin{array}{rl}\frac37&=\frac17+.\overline{285714}\\&=\frac17+\frac15+.0\overline{857142}\\&=\frac17+\frac15+\frac6{70}\\&=\frac17+\frac15+\frac1{70}+\frac1{14}\end{array}$$
Tweaking that a bit led me to$$\frac17+\frac14+\frac1{32}+\frac1{224}$$
Question: Are there more solutions? Especially: Are there infinitely many?
The greedy algorithm gives the triple $$ \frac{1}{3} + \frac{1}{11} +\frac{1}{231} $$ This leads to three quadruples using $$ \frac{1}{a} = \frac{1}{a+1} +\frac{1}{a^2 + a} $$
These would be denominator quadruples $$ (4,12,11,231) \mapsto (4,11,12,231), $$ $$ (3,12,132,231), $$ $$ (3,11,232,53292). $$ There are finitely many. We demand that denominators be written in increasing order. The first can be no larger then $4 \cdot( 7/3 ).$ If $10$ or larger, $$ 4 (1/10) = 2 / 5 < 3/7 $$ For each choice of the first denominator, there is likewise a bound on the second denominator, and so on.