# Unique solution of the equation $\prod_{i=1}^m(a_i+1)\prod_{i=1}^n(b_i)=\prod_{i=1}^m(a_i)\prod_{i=1}^n(b_i+1)$

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).

• Would it help if we write the equation in the following form and use inequality arguments? $$\prod_{i=1}^{m}\left(1+\frac{1}{a_i}\right)=\prod_{i=1}^{n}\left(1+\frac{1}{b_i}\right)$$ – Shubhrajit Bhattacharya May 23 at 10:13
• Rewriting the equation as proposed might be helpful. But how we may use inequality arguments? – Eldar Sultanow May 23 at 10:49
• 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? – Eldar Sultanow May 23 at 11:07

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 $$(b_1,b_2)=(6335,17919)$$:
$$4888\cdot110592\cdot6335\cdot17919=4887\cdot110591\cdot6336\cdot17920$$