Minimum number of fractions to be summed up to $\frac45$ What is the minimum number of fractions having numerator 1 and a natural number as denominator to be summed up to $\frac 45$?
I have tested with 2 fractions: $\frac1a + \frac1b = \frac45$ and get into the diophantine  equation: $5(a+b)=4ab$ and it seems this should have some solutions but can't find one!!

Considering prime factorization gives that $5 \mid ab$ and, by relabeling if necessary, we can assume that $5 \mid a$, and in particular $\frac{1}{a} \leq \frac{1}{5}$. Substituting in the two-fraction equation gives $b \leq \frac{5}{3}$, so we must have $b = 1$, but this gives $\frac{1}{a} + 1 = \frac{4}{5}$, and the solution $a$ to this equation is negative, so there is no decomposition of $\frac{4}{5}$ as a sum of two such fractions.
On the other hand, if we permit ourselves three such fractions, we can see immediately that one of the fractions must be $\frac{1}{2}$ or $\frac{1}{3}$. (If not, each of the three fractions would be $\leq \frac{1}{4}$ and so would have sum $\leq 3 \cdot \frac{1}{4} < \frac{4}{5}$.) So, this reduces the problem to finding decompositions of either $\frac{3}{10}$ or $\frac{7}{15}$ into two such fractions. There turn out to be two solutions. (Incidentally, fractions of the form $\frac{1}{n}$, where $n$ is a positive integer, are often called Egyptian fractions.)
