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](https://doi.org/10.1007/978-3-319-68032-3_20) 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](http://maths.nju.edu.cn/~zwsun/UnitFraction.pdf) 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 $r<1$.



Can we write each positive rational number as $\frac1{p_1-1}+\ldots+\frac1{p_k-1}$ with $p_1,\ldots,p_k$ distinct primes?
This conjecture is true (as is the version for $p-h$ for any $h\neq 0$).


The proof is too long to reproduce here, but the preprint is at <https://arxiv.org/abs/2305.02689>


EDIT:


Quick summary as requested: My previous paper (<https://arxiv.org/abs/2112.03726>) essentially shows that for any set $A$ which is large enough and satisfies some mild number theoretic conditions we can write $1$ as the sum of distinct unit fractions with denominators from $A$. (The most important of these conditions is 'smoothness', that no $n\in A$ is divisible by a large prime power.)


Such a result was previously proved by Croot in 2003, but with a stronger smoothness condition. The proof is a strengthening of Croot's method, using the circle method and some combinatorial analysis.


What's new is observing that we can show that the set $\{p-h\}$ satisfies these mild number theoretic conditions - which is to be expected, since these conditions are all 'generic', and we expect $p-h$ to look like a typical integer of the same size from a multiplicative point of view. In fact very classical techniques suffice here (e.g. the Siegel-Walfisz theorem, Selberg's sieve, and an Erdos-Kac type theorem due to Halberstam from the 1950s).


This means we can write $1$ as the sum of $\frac{1}{p-h}$. But this is robust, in that we can remove any finite set of primes and still find such a sum. Using the identity $n=1+\cdots+1$ we can therefore write any integer $n$ as the sum of distinct $\frac{1}{p-h}$. Now to write any $\frac{n}{m}$ do the above with primes $p\equiv h\pmod{m}$ and the set $A=\{\frac{p-h}{m}\}$.


