Let, $a_i, i=1,2, \ldots, n$ be distinct real numbers $b_1 b_2, \ldots, b_n$ be real numbers, such that the product $\prod_{j=1}^n\left(a_i+b_j\right)$ is the same for each $i$. Prove that the product $\prod_{i=1}^n\left(a_i+b_j\right)$ is also constant for all $j$.
I am getting some issues here. It's given $a_i$ are distinct, so WLOG we can assume $a_1 <a_2<a_3< \dots <a_n$ and that would make $(a_1+b_1) < (a_2+b_1)$ and so on for all $a_i$, but as we are given in the problem, $\prod_{j=1}^n\left(a_i+b_j\right)$ is same for all $i$, we get $(a_1+b_1)(a_1+b_2) \dots (a_1+b_n)= (a_2+b_1)(a_2+b_2) \dots (a_2+b_n)$. It's contradicting. Where am I going wrong?
Let $Q(x)$ be the polynomial defined $$Q(x)=-C+\prod_{j=1}^n (x+b_j),$$ where $C$ is defined to be the constant $$C=\prod_{j=1}^n (a_1+b_j).$$
Then, from the hypothesis, the equation $Q(a_i)=0$ follows for each $i=1,2,\ldots,n$. Furthermore, as $Q$ has degree $n$ and the $a_i$s are all distinct, it follows that the roots of $Q$ is precisely the set $\{a_1,a_2,\ldots,a_n\}$, so $Q(x)$ can be rewritten: $$Q(x)=\prod_{i=1}^n (x-a_i).$$ Thus, the numbers $\prod_{i=1}^n (a_i+b_j)$; $j=1,2,\ldots,n$ are all the same if $\prod_{i=1}^n (-b_j-a_i)$ $=$ $Q(-b_j)$ $=$ $(-1)^n\prod_{i=1}^n (a_i+b_j)$ is the same for all $j=1,2,\ldots,n$.
However, indeed [looking at how $Q(x)$ was originally defined]: $$\forall j=1,2,\ldots, n:$$ $$Q(-b_j) =-C+ \prod_{i=1}^n (b_i-b_j)$$ $$=-C + 0 = -C.$$ So indeed, $Q(-b_j)$ is indeed the same for all $j=1,2,\ldots,n$, and so as noted above in the bold, the result follows.