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 Tau of Power of Prime, $\map \tau {p^k} = k + 1 \ge 9$. Therefore we have $k \ge 8$. {{qed|lemma}} === 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 $C$. {{qed|lemma}} ==== 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 $C$. {{qed|lemma}} Cases $1$ and $2$ above shows that for each prime $p$ dividing $C$, we must have $p^3$ dividing $C$ as well. {{qed|lemma}} === 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 $2$: :$C = p_1^{a_1} p_2^{a_2} k \ge 2^{a_1} 3^{a_2} \ge 2^3 3^3 = 216$ {{qed}} {{explain|There are a number of assumptions which appear to made for the purposes of a Proof by Contradiction -- worth labelling them as such and indicating where those assumptions are refuted, for clarity.}}