It is well-known that any positive rational number can be written as the sum of finitely many distinct unit fractions. This is easy since $$\frac1n=\frac1{n+1}+\frac1{n(n+1)}\quad\text{for all}\ n=1,2,3,\ldots.$$ In 2015 I thought that this easy fact should have a further refinement which is somewhat sophisticated. Note that the series $\sum_p\frac1{p-1}$ and $\sum_p\frac1{p+1}$ (with $p$ prime) diverge just like the harmonic series $\sum_{n=1}^\infty\frac1n$. Also, for any positive integer m there are infinitely many primes $p$ congruent to $1$ (or $-1$) modulo $m$ (by Dirichlet's theorem). Motivated by this, I made the following conjecture in Sept. 2015.

**Conjecture**. For any rational number $r>0$, there are finite sets $P_r^-$ and $P_r^+$ of primes such that
$$r=\sum_{p\in P_r^-}\frac1{p-1}=\sum_{p\in P_r^+}\frac1{p+1}.$$

This appeared as Conjecture 4.1 of this published paper of mine. For example, $$2=\frac1{2-1}+\frac1{3-1}+\frac1{5-1}+\frac1{7-1}+\frac1{13-1}$$ with $2,3,5,7,13$ all prime, and $$1=\frac1{2+1}+\frac1{3+1}+\frac1{5+1}+\frac1{7+1}+\frac1{11+1}+\frac1{23+1}$$ with $2,3,5,7,11,23$ all prime. Also, \begin{align*}\frac{10}{11}=&\frac1{3-1}+\frac1{5-1}+\frac1{13-1}+\frac1{19-1}+\frac1{67-1}+\frac1{199-1} \\=&\frac1{2+1}+\frac1{3+1}+\frac1{5+1}+\frac1{7+1}+\frac1{43+1}+\frac1{131+1}+\frac1{263+1} \end{align*} with $2,3,5,7,13,19,43,67,131,199,263$ all prime. The reader may see more numerical data in my detailed introduction to this conjecture.

After learning this conjecture from me, Prof. Qing-Hu Hou and Guo-Niu Han checked my above conjecture seriously and their computational results support my conjecture. For example, in 2018 Prof. Han found 2065 distinct primes $p_1<\ldots<p_{2065}$ with $p_{2065}\approx 4.7\times10^{218}$ such that $$\frac1{p_1+1}+\ldots+\frac1{p_{2065}+1}=2.$$

My question is whether the above conjecture is true. I would like to offer 500 US dollars as the prize for the first correct solution.

**Remark**. Let $r$ be any positive rational number, and let $\varepsilon\in\{\pm1\}$. As the series $\sum_p\frac1{p+\varepsilon}$ (with $p$ prime) diverges, there is a unique prime $q$ such that $$\sum_{p<q}\frac{1}{p+\varepsilon}\le r<\sum_{p\le q}\frac1{p+\varepsilon}.$$ Thus
$$0\le r_0:=r-\sum_{p<q}\frac1{p+\varepsilon}<\frac1{q+\varepsilon}\le1.$$
If $r_0=\sum_{j=1}^k\frac1{p_j+\varepsilon}$ with $p_1,\ldots,p_k$ distinct primes, then $p_1,\ldots,p_k$ are all greater than $q$, and
$$r=\sum_{p<q}\frac1{p+\varepsilon}+\sum_{j=1}^k\frac1{p_j+\varepsilon}.$$ Therefore it suffices to consider the conjecture only for $r<1$.

Egyptianif every positive rational number admits an Egyptian fraction decomposition with all the denominators in $S$, and then ask for sufficient conditions for $S$ to be Egyptian. Perhaps it suffices for $S$ to have high enough density and to satisfy certain congruence conditions? $\endgroup$ – Timothy Chow Nov 29 '18 at 21:37