# 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?

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 $$p_1<\ldots with $$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 $$\sum_{p Thus $$0\le r_0:=r-\sum_{p 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 $$r=\sum_{p Therefore it suffices to consider the conjecture only for $$r<1$$.

• Voting for this question because it rekindles in me the sheer joy of natural numbers. There will, I sincerely hope, always be people interested in 'strange' questions like this one, that provoke the creative mind. Nov 29, 2018 at 10:38
• Interesting conjecture. One could consider defining a set $S$ of natural numbers to be Egyptian if 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? Nov 29, 2018 at 21:37

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 $$A=\{\frac{p-h}{m}\}$$.

• What about a summary? May 5 at 10:47
• Added a summary May 5 at 10:54
• Many thanks -- it's so nice! May 5 at 11:09
• Dr. Thomas, thank you very much for your surprising proof! May 6 at 7:32