Representation of a number as a product of $\sqrt{n^2 + 1} + n$

Question

Question. Do there exist two multisets $A, B$ consisting of positive integer numbers such that $|A|$ and $|B|$ have different parity and $$ \prod_{n\in A}(n + \sqrt{n^2 + 1}) = \prod_{m\in B}(m + \sqrt{m^2 + 1})? $$

Thoughts. The parity condition is essential, e.g., $(1 + \sqrt{2})^3 = 7 + 5\sqrt{2} = 7 + \sqrt{7^2 + 1}$.

If $\sqrt{n^2 + 1} = m\sqrt{d}$ with square-free $d$ then $n^2 - dm^2 = -1$, i.e., $(n, m)$ is a solution of a negative Pell's equation and it can be represented as $$ n + \sqrt{n^2 + 1} = n + m\sqrt{d} = (n_d + m_d\sqrt{d})^{2k + 1}, $$ where $(n_d, m_d)$ is a fundamental solution of Pell's equation and $k$ is a non negative integer. This means that in the initial equation one can consider only multipliers of the form $n_d + \sqrt{n_d^2 + 1}$.

Almost equivalent question. Let $D$ be a set of integers $d$ such that the negative Pell's equation $n^2 - dm^2 = -1$ is solvable. Can a real number $x$ have two different representations of the form $$ x = \prod_{d_i\in D}(n_{d_i} + m_{d_i}\sqrt{d})^{\alpha_i}, \qquad \alpha_i\in \mathbb{Z}_+. $$

Answer 1

$\def\supp{\mathop{\mathrm{supp}}}$ Surely, in the ``Almost equivalent question'' you assume that the $d$'s are square-free.

Denote $R=\mathbb Z[\sqrt{p_1},\dots,\sqrt{p_n}]$. We prove that in its group of units $U_R$, the elements $u_d=n_d+m_d\sqrt{d}$ (for all possible $d$) are independent (even modulo roots of unity $\pm1$), thus establishing a negative answer to both questions.

For that purpose, implement a proof of Dirichlet's unit theorem (see, e.g., p. 87 here). $R$ has $2^n$ automorphisms indexed by $I=(i_1,\dots,i_n)\in C=\{0,1\}^n$ and given by $\sigma_I(\sqrt{p_k})=(-1)^{i_k}\sqrt{p_k}$. Consider the group homomorphism $L\colon U_R\to\mathbb R^{C}$ given by $L(u)_I=|\ln \sigma_I(u)|$. It suffices to show that all the $v_d=L(u_d)$ are linearly independent. We show that they are linearly independent along with $v_0$ all whose coordinates are ones.

If $d=\prod_{j\in J}p_j$, then $L(u_d)_I$ attains a constant for all $I$ with $|J\cap \supp I|$ even, and another constant for all other $I$. So, modulo $v_0$, we may replace $v_d$ with $w_J$ such that $(w_J)_I=(-1)^{|J\cap \supp I|}$. All such $w_J$ (for all $J$) are well known to be independent (in fact, they consist just of the values of $\prod_{j\in J}(1-2x_j)$ on $C$).

NB. In fact, we have proved a more general fact. If, for any $\varnothing\neq J\subseteq \{1,\dots,n\}$ we choose a number $$ u_J=n_J+m_J\sqrt{\prod_{j\in J} p_j}, $$ where $n_J,m_J$ are nonzero rational numbers, and we also choose $u_\varnothing \in \mathbb Q\setminus\{0!\pm1\}$, then all products of the chosen numbers (with integer exponents) are distinct.