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Stated simply the question is: given the set of an arithmetic sequence of cardinality $2N$, where $N$ is greater than or equal to $2$, is it possible to choose $N$ integers in such a way that their respective products is equal.

Example: If choose a set of size $4$ with numbers $1, 2, 3$ and $6$ (which I won't because the numbers don't have a common difference), i would get: $2*3$ = $1*6$.

If this is possible what is there any formula, that bypasses the need for the divisor function's inclusion?

It would be really interesting as this question can be modified to ask different question which perhaps ask the deepest question of relation between sums and products.

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  • $\begingroup$ @PeterMueller The numbers should be positive integers. $\endgroup$
    – Stan
    Commented Nov 25 at 13:30

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A necessary condition for an example is that there are positive integers $a,b$ such that $\prod_{i=0}^{2N-1}(a+ib)$ is a square. Differently phrased (divide by $b^{2N}$), we ask for rational solutions of the hyperelliptic curve $Y^2=\prod_{i=0}^{2N-1}(X+i)$. According to a conjecture by Sander (see Nontrivial rational points on Erdős-Selfridge curves), there are no solutions except a trivial one which does not meet your positivity requirement.

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