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It is easy to see that if $\prod_{i=1}^n(1-x^{k_i})=\prod_{i=1}^n(1-x^{l_i})$ then $k_i=l_j$ for some $i$ and $j$.

Let $m_i,n_i, (i=1,2, \cdots, 8)$ be positive integers such that

$$[(1-x^{m_1})(1-y^{m_2})+(1-x^{m_2})(1-y^{m_1})][(1-x^{m_3})(1-y^{m_4})+(1-x^{m_4})(1-y^{m_3})](1-x^{m_5})(1-y^{m_6})(1-x^{m_7})(1-y^{m_8})=[(1-x^{n_1})(1-y^{n_2})+(1-x^{n_2})(1-y^{n_1})][(1-x^{n_3})(1-y^{n_4})+(1-x^{n_4})(1-y^{n_3})](1-x^{n_5})(1-y^{n_6})(1-x^{n_7})(1-y^{n_8}).$$ Is it true that $\{m_1,m_2\}=\{n_1,n_2\}$ or $\{n_3,n_4\}$ ?

By putting $x=y$ we conclude (from the above result) that the set $\{m_i:i=1,2, \cdots, 8\} $ is same as the set $\{n_i:i=1,2, \cdots, 8\}$. I believe the above statement is true as well.

user26857
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jack
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    Your formula probably has a typo since the two terms in the first sum are the same. What’s the context for this? – Eric Jul 10 '21 at 18:42
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    Thanks a lot for pointing out the typo. I believe similar statement in two variables (like the one in first line) must be true if we replace some of the terms by permuting $x$ and $y$. For just two products on both sides by brute force I could prove that the statement is true. In this case by comparing the monomials I get many conditions on $m_i$'s and $n_i$'s but nothing is conclusive. There may be a smart way to reach the conclusion. – jack Jul 10 '21 at 19:11

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This seems to be a trivial class of counterexamples, but it’s not clear how you’d want to amend your conjecture without more context (there seem to be several different options that come to mind).

$$ m_1 = m_2 = n_7 = n_8 = a, \\ m_7 = m_8 = n_1 = n_2 = b, $$ and $m_i = n_i$ arbitrarily for $3 \le i \le 6$.

Then both sides, ignoring the identical factors from $3 \le i\le 6$, multiply out to $2(1-x^a)(1-y^a)(1-x^b)(1-y^b)$, but $\{m_1, m_2\}$ is not equal to $\{n_1,n_2\}$ nor $\{ n_3,n_4\}$.

Erick Wong
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