Product Grid

No reason. The question asked whether it was possible. I showed it was.

I think this works. Put $x_{i,j} \; = \; A^uB^v \hspace{2cm} 1 \le I,j \le N$ where $1 \le u,v \le N$ are the uniquely chosen integers such that $i - j \; \equiv \; u \hspace{1cm} i + j \; \equiv \; v \hspace{1cm} \pmod{N}$ and $A,B$ are coprime positive integers. The common row/column product is then $(AB)^{\frac12N(N+1)}$ . This is because the elements of each row/column contain the factors $A^j$ , $B^k$ for $1 \le j,k \le N$ multiplied together in $AB$ -pairs in some order.

This works for $N$ odd, but that is OK for $N=2017$ .

1st cheap variation: (of Mark Hennings solution): consider a Graeco-Latin square (see https://en.wikipedia.org/wiki/Graeco-Latin_square ). Associate to the corredponding Roman and Greek letters the same number (e.g. $a$ and $\alpha \to 1$ , $b$ and $\beta \to 2$ , etc...). Pick two coprime numbers A and B. Put the number $A^a B^\alpha$ (and so forth) in the square $a \alpha$ , etc... Graeco-Latin squares are known to exist for any number $\geq 3$ except 6, but I honestly don't know how exlpicit the proof is.

2nd cheap variation: take any solutions (i.e. $p$ as base and the magic square entries as exponents, or A and B as base and the Graeco-Latin square entries as exponent). As long as the bases of the exponents are all coprime, you can multiply square-wise the two solutions and get a new solution.

3rd cheap variation: pick two lists of numbers $a_1, a_2, \ldots , a_{2017}$ and $b_1, b_2, \ldots , b_{2017}$ . All those numbers should be coprime to each other. instead of sending the Graeco latin pair $(x,y)$ to $A^x B^y$ , send them to $a_x a_y$ . Actually "coprime" is a bit overkill, some weaker hypothesis would do...

I have an idea which could be worked up into a solution. It does not rely on any prior knowledge of magic or Latin squares.

Let $p_n$ be the $n^{th}$ prime number. Here is how my solution would look for a three by three square -

$\begin{array}{ccc} p_1 p_4 & p_2 p_6 & p_3 p_5 \\ p_2 p_5 & p_3 p_4 & p_1 p_6 \\ p_3 p_6 & p_1 p_5 & p_2 p_4 \end{array}$

Each row and each column contain the first six prime numbers and so they all have the same magic product $2 \times 3 \times 5 \times 7 \times 11 \times 13$ . Moreover, by the fundamental theorem of arithmetic, all of the nine entries in the square are different.

Explicitly; For the 2017 by 2017 square the first multiplicands in each column cycle through the first 2017 prime numbers starting with the first prime number in the first place in the first column, the second prime number in the first place in the second column and so on. The second multiplicands in each column cycle backwards from the $4034^{th}$ to the $2018^{th}$ prime numbers starting from the bottom of the column and again changing the starting point for each new column as shown in the three by three example.

For this construction we get the (ridiculously large) magic product

$\prod_{1}^{4034}p_i$

True, but if the elements of a magic square are the indices, to the same base, of the entries in our square, then the products of the elements in the rows and columns (and the main diagonals) of that new square are all the same, because the index sums are all the same.

Taking the exponent of sums gives products. If

is a magic square with equal sums, then taking 2 to the power of them:

is a magic square with equal products.

By taking powers, sums of the exponents become products of the powers.