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

Solution 1:

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 $\prod (a_k^2+1)$)

Solution 2:

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 $(3)$.

Solution 3:

Hint: Just translating your language a little bit, consider $A=\{a_1,\cdots ,a_n\}$ and consider a weight on a set $q(S)=\prod _{s\in S}a_s$ (denote $q(\emptyset)=1$, that is where the $2$ is coming.) Here $u:\mathcal{P}(S_n)\longrightarrow \mathbb{Z}[a_1,\cdots a_n,\frac{1}{a_1},\cdots a_n]$ is defined as $u(S)=q(S)+q(S)^{-1}.$

Show that $$\prod _{i=1}^n u(\{i\})=\sum _{\substack{A\cup B=S_n\\A\cap B=\emptyset}}\frac{q(A)}{q(B)}.$$ here an induction argument suffices.

Show that $q(X)=q(A)q(B)$ if $A\cup B=X$ with $A\cap B=\emptyset.$

Notice that $A,B$ gives you two summands in the RHS of the equation above.