TITLE: Can we arrange $\{1,...,16\}$ in $4\times 4$-grid so {products of rows} = {products of columns}?
QUESTION [14 upvotes]: My motivation is the following question asked by Jacob Steiner, which he then deleted.  For which $n\in\Bbb N$ is it possible to arrange $\{1,\ldots,n^2\}$ in an $n\times n$-grid so that the set of products of columns equals the set of products of rows?
The answer for $n=2$ is clearly No, since the only possibility, up to a transposition and a permutation of rows and columns, is
1 2 
3 4 

Since the set of products of rows are $\{2,12\}$ and the set of products of columns are $\{3,8\}$, this arrangement does not work.  So for $n=2$, it is not possible to make such an arrangement.

REPLY [2 votes]: Not an answer, but pointing out a flaw in the OP logic.
If a prime factor appears singly (i.e. not squared) an odd number of times (like $5$ appearing as $\{5, 10, 15\}$ in the $4\times 4$ grid), that does not imply one of the appearances must be along the diagonal.  E.g.
* 10  *  *
*  * 15  *
5  *  *  *
*  *  *  *

would meet the requirement as far as factors of $5$ are concerned: the first $3$ rows and the first $3$ columns each has exactly one factor of $5$ (and the last row and last column has no factor of $5$).
(These positions represent a (fixed-point-free) permutation in the $3 \times 3$ submatrix.)
UPDATE: Indeed Rob Pratt found such a matrix:
$$
\begin{matrix}   9 &15 &12  &1\\   3 &11 &14  &5\\  6  &7 &13 &16\\  10 &2  &4  &8\end{matrix}
$$
where the positions of the multiples of $5$ represent a fixed-point-free permutation of the $3\times 3$ submatrix after deleting the $3$rd row & column.
So the OP claim that the $4\times 4$ grid must have a factor of $5$ along the diagonal is wrong, and similarly, it is still unproven that $8 \times 8$ cannot be filled because that last $X$ does not need to be any multiple of $11, 17, 19$.