Question 166
Let $a_{1}, a_{2}, \ldots, a_{n}$, and $b_{1}, b_{2}, \ldots, b_{n}$ be real numbers, and let $a_{1}, a_{2}, \ldots, a_{n}$ be all different. Show that if all the products
$$
\left(a_{i}+b_{1}\right)\left(a_{i}+b_{2}\right) \cdots\left(a_{i}+b_{n}\right)
$$
for $i=1,2, \ldots, n$, are equal, then the products
$$
\left(a_{1}+b_{j}\right)\left(a_{2}+b_{j}\right) \cdots\left(a_{n}+b_{j}\right)
$$
for $j=1,2, \ldots, n$, are equal, too.
$$
\left(a_{i}+b_{1}\right)\left(a_{i}+b_{2}\right) \cdots\left(a_{i}+b_{n}\right)
$$
for $i=1,2, \ldots, n$, are equal, then the products
$$
\left(a_{1}+b_{j}\right)\left(a_{2}+b_{j}\right) \cdots\left(a_{n}+b_{j}\right)
$$
for $j=1,2, \ldots, n$, are equal, too.