$\prod_{I = 1}^N(x + y_i) = \prod_{I = 1}^N(x - y_i)^{\alpha_i}$ where all elements are non-zero natural numbers and $x > \forall y_i$ for any $x$?

Question

I am interested if there are possible criteria one can place on the $y_i$'s which will give solutions to this relationship. The simplest way I can think of is to allow some positive integer $m_i, n_i$ where $x - y_i = n_i$ and $x + y_i = m_i$ allowing for $(x, y_i) = (\frac{m_i + n_i}{2}, \frac{m_i - n_i}{2})$ where $2x = m_i + n_i$ for all $i$. It is clear that when $GCD(2x, n_i) = d_i$, then $d_i|m_i$, but do all non trivial solutions require $GCD(2x, n_i) > 1$?

Answer 1

If there is only one term in the product, it becomes $$x+y=(x-y)^{\alpha}.\tag{1}$$ Set $u:=x+y$ and $v:=x-y$ so that $u\equiv v\pmod{2}$ and $u>v$. We recover $x$ and $y$ by $x=\tfrac{u+v}{2}$ and $y:=\tfrac{u-v}{2}$. Then the above becomes $$u=v^{\alpha},$$ where clearly $u\equiv v\pmod{2}$, and we have $u>v$ if and only if $v,\alpha>1$. This shows that all solutions to $(1)$ are given by $$(x,y,\alpha)=\left(\frac{v^{\alpha}+v}{2},\frac{v^{\alpha}-v}{2},\alpha\right),\qquad v,\alpha>1.$$