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Let $a_i$ be a sequence of $m$ distinct odd integers and $b_i$ a sequence of $n$ distinct odd integers.

We have to prove that, $$\prod_{i=1}^m(a_i+1)\prod_{i=1}^n(b_i)=\prod_{i=1}^m(a_i)\prod_{i=1}^n(b_i+1)$$ has only one solution: $n=m$ with $a_i=b_i$ for $1\le i\le n$ (neglecting the sorting of sequence members).

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    $\begingroup$ Would it help if we write the equation in the following form and use inequality arguments? $$\prod_{i=1}^{m}\left(1+\frac{1}{a_i}\right)=\prod_{i=1}^{n}\left(1+\frac{1}{b_i}\right)$$ $\endgroup$ – Shubhrajit Bhattacharya May 23 at 10:13
  • $\begingroup$ Rewriting the equation as proposed might be helpful. But how we may use inequality arguments? $\endgroup$ – Eldar Sultanow May 23 at 10:49
  • $\begingroup$ Maybe we can rewrite $\prod_{i=1}^m(1+a_i)=\sum_{|\alpha|=0}^m\prod_{k=1}^ma_k^{\alpha_k}$ whereby $\alpha$ is a vector of $m$ binary elements $\alpha\in\{0,1\}^m$, would this be helpful? $\endgroup$ – Eldar Sultanow May 23 at 11:07
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For a disproof a counterexample is sufficient: The both sequences $(a_1,a_2,a_3)=(135,85,215)$ and $(b_1,b_2,b_3)=(65,165,415)$ satisfy the introduced product equality:

$136\cdot86\cdot216\cdot65\cdot165\cdot415=135\cdot85\cdot215\cdot66\cdot166\cdot416$

Another example are the sequences $(a_1,a_2)=(4887,110591)$ and $(b_1,b_2)=(6335,17919)$:

$4888\cdot110592\cdot6335\cdot17919=4887\cdot110591\cdot6336\cdot17920$

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