Sets of integers with same sum and same sum of reciprocals

Question

Let $A$, $B$ be two distinct sets of natural numbers. Is it possible to have $\ \sum_{a\in A}a\,=\,\sum_{b\in B}b\ $ and $\ \sum_{a\in A}1/a\,=\,\sum_{b\in B}1/b\ $ at the same time?

Answer 1

Yes. Take a large $N$. Let $E := \{n \le N : p \mid n \implies p \le \frac{N}{100}\}$, where $p$ stands for a prime.

Note that $$\biggl|[N]\setminus E\biggr| \le \sum_{\frac{N}{100} \le p < N} \frac{N}{p} \le C\frac{N}{\log N},$$ implying \begin{equation}\tag{1} |E| \ge \Bigl(1-o(1)\Bigr) N.\end{equation}

Now, $$\log \text{lcm}(E) = \log \prod_{p \le \frac{N}{100}} p^{\lfloor \log_p N \rfloor} \le \sum_{p \le \frac{N}{100}} \log N \le \frac{N}{10}.$$

In other words, \begin{equation}\tag{2} \text{lcm}(E) \le e^{N/10} \le 2^{N/5}.\end{equation}

Therefore, for any $A \subseteq E$, we have $\sum_{n \in A} \frac{1}{n} = \frac{p}{\text{lcm}(E)}$ for some $p \le N \, \text{lcm}(E) \le 2^{N/4}$.

Finishing up, each $A \subseteq E$ gives rise to a pair $(\sum_{n \in A} n, \text{lcm}(E)\sum_{n \in A} \frac{1}{n})$, which lives in $[N^2] \times [2^{N/4}]$, the size of which is at most $2^{N/3}$. Since there are $2^{|E|} \ge 2^{N/2}$ many subsets of $E$, the pigeonhole principle guarantees two (distinct) with the same pair.

Answer 2

For any non-zero $(a, b, c)$ with $b^2\ne ac$, take $(x/c, x/b, x/a)$ where $(1/a+1/b+1/c)x=a+b+c$. If the resulting numbers aren't all integers, then multiply all six by a common factor to clear fractions.

Answer 3

A simplified version of mathworker21's argument.

Denote $N=6^{T-1}$, then the set $A$ of divisors of $N$ has $|A|=\tau(N)=T^2$ elements. Consider a subset $B$ of $A$, there are $2^{T^2}$ choices of $B$. Denote $w(B):=$(sum of elements $B$, sum of reciprocals of elements of $B$). The sum of elements of $B$ does not exceed $\sigma(N)=(1+2+\ldots+2^{T-1})(1+3+\ldots+3^{T-1})<2^T\cdot 3^T=6^T$. The sum of their reciprocal multiplied by $N$ is an integer and does not exceed the same value $6^T$. Thus, the number of distinct pairs $w(B)$ does not exceed $6^T\cdot 6^T=36^T$ that is much less than $2^{T^2}$, therefore, there exist many subsets with the same $w(B)$.

Answer 4

Here is a 4-parameter family of solutions with $|A|=|B|=3$.

Consider the optic equation $\frac1a+\frac1b=\frac1c$; it is known that the positive integer solutions to this equation are parametrised by $$\begin{align} a&=km(m+n),\\ b&=kn(m+n),\\ c&=kmn, \end{align}$$ with $k,m,n\in\mathbb Z^+$. Hence the set $\{a,b,-c\}$ has sum $k(m^2+mn+n^2)$ and sum of reciprocals 0.

Now for any $m,n,m',n'\in\mathbb Z^+$, let $$\frac{m^2+mn+n^2}{m'^2+m'n'+n'^2}=\frac{k'}k.$$ Define $a,b,c$ as above, and $a',b',c'$ analogously. Then $$a+b-c=k(m^2+mn+n^2)=k'(m'^2+m'n'+n'^2)=a'+b'-c',$$ and $$\frac1a+\frac1b-\frac1c=0=\frac1{a'}+\frac1{b'}-\frac1{c'},$$ so the sets $A=\{a,b,c'\}$ and $B=\{a',b',c\}$ satisfy the desired conditions.

Answer 5

If one has a solution in rational numbers one can scale up to get integers.

Rather than starting with equal sums of reciprocals. one could start with two equations such as $1+5+10=3+13$ and $2+4+15=21.$ Then for any $x,y$ the sets $A=\{x,5x,10x,21y\}$ and $B=\{3x,13x,2y,4y,15y\}$ have equal sum. Setting the sums of the reciprocals equal fixes the ratio $\frac{x}{y}=r$ for some rational number $r=\frac{P}{Q}.$ Set $x=P,y=Q$ for an integer solution. If the elements have a relatively small least common multiple (so if they come from the divisors of an integer with small prime divisors). The final scaled up integer solution will have correspondingly smaller members. parametric solutions are possible, which might not differ from what has already been suggested.

If there are two solutions $(A,B)$ and $(A',B')$ then $(A\cup A',B\cup B')$ is a solutions So one might want primitive solutions which do not split this way. Of course if $A \cap B' \ne \emptyset$ and/or $A' \cap B \ne \emptyset$ then the common elements can be removed and that might give a primitive solution.