as Fibonacci-Sylvester algorithm) shows there is a solution. Indeed, let $n=my+r$ with $0 <r<m$, then $\frac{m}{n}-\frac{1}{y+1}=\frac{m-r}{n(y+1)}$ has a smaller numerator, and inductively a solution with $k {\leqslant}m$ is constructed. For an alternative method (especially if $m=k=4$) see also Schinzel [@schinzel:1956].

If $m>k{\geqslant}3$, and the $t_i$ are positive integers, then it is an open problem if for each sufficiently large $n$ there is at least one solution. The Erdős-Straus conjecture with $m=4, k=3$, discussed above, is the most prominent case. If $m$ and $k$ are fixed, one can again establish sets of residue classes, such that is generally soluble if $n$ is in any of these residue classes.

The problem of classifying solutions of has been studied by Rav [@rav], Sós [@Sos:1905] and Elsholtz [@els]. Moreover Viola [@viola], Shen [@shen] and Elsholtz [@elsholtz:2001] have used a suitable subset of these solutions to give (for fixed $m > k{\geqslant}3$) quantitive bounds on the number of those integers $n {\leqslant}N$, for which does not have any solution.

In order to study, whether there is at least one solution, it is again sufficient to concentrate on prime denominators. The average number of solutions is smaller when averaging over the primes only, but we intend 