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What is the maximal number of solutions of the following equation?

$\sum_{i = 1}^n 1/a_i^x - \sum_{i = 1}^m 1/b_i^x = 0$

where $x$ is the unknown and $n, m$, $a_i$'s, $b_i$'s are constant.

It satisfies that $x > 0$ and $a_i, b_i > 1$ for all $i$.

IN OTHER WORDS:

For which positive integer non-decreasing sequences $\ (a_1\ \ldots\ a_n)\ \ne\ (b_1\ \ldots\ b_m)\ $ the number of positive integers $\ x\ $ such that

$$\sum_{i = 1}^n \frac 1{a_i^x}\ -\ \sum_{i = 1}^m \frac 1{b_i^x}\ \,=\,\ 0$$

is maximal (and what would be that maximal number of solutions $\ x\ ?$).

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    $\begingroup$ If the b_i are a permutation of the a_i, all positive x satisfy the equation. You might consider the case m=n=2 first before looking at longer tuples. $\endgroup$ May 20 '15 at 6:00
  • $\begingroup$ Thanks! a's and b's are all distinct. I didn't notice that. $\endgroup$
    – user9836
    May 20 '15 at 6:05
  • $\begingroup$ Note that any positive rationals will do instead of $1/a_i$ and $1/b_j$ since they can be uniformly scaled into reciprocals of integers. I'm predicting there will be no limit on the number of solutions, since I don't think any number of leading coefficients of a polynomial with positive coefficients, together with knowing that all the roots are negative and rational, is enough to determine the polynomial. $\endgroup$ May 20 '15 at 8:07
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For any set $S$ let $\mu_x(S)=\sum_{s\in S} s^x$. When $S$ is a nonempty subset of $\lbrace 1,\ldots,k\rbrace$, $1\le \mu_x(S)\le k^{x+1}$, so the number of possible values of $(\mu_1(S),\mu_2(S),\ldots,\mu_t(S))$ is at most $k^2\times\cdots\times k^{t+1}= k^{t(t+3)/2}$.

Since there are $2^k-1$ non-empty subsets of $\lbrace 1,\ldots,k\rbrace$, some pigeons tell us that when $2^k-1 > k^{t(t+3)/2}$ there are two different nonempty subsets $S,S'$ of $\lbrace 1,\ldots,k\rbrace$ such that $(\mu_1(S),\mu_2(S),\ldots,\mu_t(S))=(\mu_1(S'),\mu_2(S'),\ldots,\mu_t(S'))$. By deleting common elements of $S$ and $S'$ we can assume they are disjoint and have at most $k$ elements altogether.

Now $\sum_{s\in S} s^x=\sum_{s\in S'} s^x$ implies $\sum_{s\in S} (s/k!)^x=\sum_{s\in S'} (s/k!)^x$ and $k!/s$ is an integer for $s\in \lbrace 1,\ldots,k\rbrace$. Therefore, if $2^k-1 > k^{t(t+3)/2}$, there are disjoint integer sequences $(a_1,\ldots,a_m), (b_1,\ldots,b_n)$ with $m+n\le k$ such that $x=1,2,\ldots,t$ are all solutions.

For any $t$, the inequality $2^k-1 > k^{t(t+3)/2}$ holds for large enough $k$, approximately $k\gt t^2\log_2 t$.

This argument can be tightened in various ways. Solutions with $m=n$ need only slightly larger $k$.

To get sharper bounds, note that for random subsets $S$, $(\mu_1(S),\mu_2(S),\ldots,\mu_t(S))$ concentrates near its ($t$-dimensional) mean. This implies that some box gets two pigeons rather sooner than the simple calculation gave.

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