Smallest Magic Constant of Order 3 Multiplicative Magic Square

Theorem
The magic constant of the smallest possible multiplicative magic square with the smallest magic constant is as follows.

Proof
From Smallest Multiplicative Magic Square is of Order 3, the smallest possible multiplicative magic square is of order $3$.

Let $C$ be a magic constant for an order $3$ multiplicative magic square.

We prove that $C \ge 216$ is smallest in $2$ steps:


 * Step $1$: If $C$ is a prime power $p^k$, then $k \ge 8$.
 * Step $2$: For each prime $p$ dividing $C$, we must have $p^3$ dividing $C$ as well.

Then $216 = 2^3 \cdot 3^3$ is the smallest such constant.

Step $1$
Suppose that $C = p^k$ for some prime $p$.

Since the multiplicative magic square has distinct entries, and each entry of the magic square is a factor of $C$, $C$ must have at least $9$ distinct factors.

From Divisor Counting Function of Power of Prime, $\map \tau {p^k} = k + 1 \ge 9$.

Therefore we have $k \ge 8$.

Step $2$
Suppose on the contrary, for some prime $p$ dividing $C$, $p^3$ does not divide $C$.

Then we have either:
 * $p \divides C$ and $p^2 \nmid C$

or:
 * $p^2 \divides C$ and $p^3 \nmid C$

Case 1: $p \divides C$ and $p^2 \nmid C$
Since $p \divides C$, each row, column and diagonal contains one entry divisible by $p$.

However there cannot be anymore, since $p^2 \nmid C$.

These entries can be in the middle, at a corner or at an edge.

We will show that none of these will form a magic square.

We indicate the entries divisible by $p$ with $p$, and the entries cannot be divisible by $p$ with $*$.

Suppose an entry divisible by $p$ is in the middle.

Then:


 * $\begin{array}{|c|c|c|}

\hline * &  * &  * \\ \hline * &  p &  * \\ \hline * &  * &  * \\ \hline \end{array}$

But then the first row is not divisible by $p$, so their product is not $C$.

Suppose an entry divisible by $p$ is at an edge. Then:


 * $\begin{array}{|c|c|c|}

\hline * &    &    \\ \hline p &  * &  * \\ \hline * &    &    \\ \hline \end{array}$

By the diagonal conditions, the top right and bottom right entries must be divisible by $p$.

But then the rightmost column is divisible by $p^2$, so their product is not $C$.

Suppose an entry divisible by $p$ is in a corner.

Then:


 * $\begin{array}{|c|c|c|}

\hline p &  * &  * \\ \hline * &  * &    \\ \hline * &    &  * \\ \hline \end{array}$

But then the diagonal from bottom left to top right has no entries divisible by $p$, so their product is not $C$.

Case $2$: $p^2 \divides C$ and $p^3 \nmid C$
The cases where an entry divisible by $p^2$ exists is similar to Case $1$.

So we suppose there is no entry divisible by $p^2$.

Similar to Case 1, we indicate the entries divisible by $p$ with $p$, and the entries cannot be divisible by $p$ with $*$.

Each row must contain an entry not divisible by $p$.

Then the cases will be the inverse of the cases in Case 1, with $p$ interchanged with $*$.

For example, the cases where an entry not divisible by $p$ is at the corner.

Then:


 * $\begin{array}{|c|c|c|}

\hline * &  p &  p \\ \hline p &  p &    \\ \hline p &    &  p \\ \hline \end{array}$

But then every entry on the diagonal from bottom left to top right are divisible by $p$, so their product is not $C$.

Cases $1$ and $2$ above shows that for each prime $p$ dividing $C$, we must have $p^3$ dividing $C$ as well.

Proof that $216$ is the smallest $C$
Now suppose $C$ is a prime power.

Then from our conclusion in Step $1$:
 * $C = p^k \ge 2^k \ge 2^8 = 512$

Suppose $C$ is not a prime power.

Then $C$ must be divisible by at least $2$ primes.

Then from our conclusion in Step $2$:
 * $C = p_1^{a_1} p_2^{a_2} k \ge 2^{a_1} 3^{a_2} \ge 2^3 3^3 = 216$