A conjecture regarding products of $u(x)=x+\frac1x$

The key fact to know is $$ u(x)u(y) = (x+1/x)(y+1/y) = u(xy)+u(x/y). \tag{1} $$ Given non-zero numbers $\,a_1,a_2,\dots\,$ define the following sequences: $$ y_n := \prod_{k=1}^n a_n, \;\; P_n := \prod_{k=1}^n u(a_n), \;\; Q_n(t) := P_n\, u(t^2y_n). \tag{2} $$ It is easy to check that $$ Q_n(t) = Q_{n-1}(t) + Q_{n-1}(t\,x_n) \tag{3} $$ expands into a $\,2^n\,$ term sum with one term for each divisor of $\,y_n.\,$

Your conjecture is that $$ \Pi_n = Q_n(1) \tag{4} $$ which is true because $Q_n(t)$ is defined as a product of $\,n+1\,$ factors in equation $(2)$ and expands into a $\,2^n\,$ term sum via equation $(3)$.

Maybe we could break the product into two expression: $$\Pi_n:=u\left(\prod_{K=1}^n a_k\right)\prod_{K=1}^n u(a_k)$$ $$=\left(\prod_{K=1}^n a_k\right)\left(\prod_{K=1}^n u(a_k)\right)+\left(\prod_{K=1}^n \frac{1}{a_k}\right)\left(\prod_{K=1}^n u(a_k)\right)$$ $$=\prod_{K=1}^n a_k\cdot u(a_k)+\prod_{K=1}^n \frac{1}{a_k}\cdot u(a_k)$$ $$=\prod_{K=1}^n (a_k^2+1)+\prod_{K=1}^n \left(\frac{1}{a_k^2}+1\right)$$ Now try to rewrite $\prod (a_k^2+1)$ and $\prod\left(\frac{1}{a_k^2}+1\right)$ as sums (you already did it for $\prod (a_k^2+1)$)