Smallest Magic Constant of Order 3 Multiplicative Magic Square

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Theorem

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

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

\hline 18 & 1 & 12 \\ \hline 4 & 6 & 9 \\ \hline 3 & 36 & 2 \\ \hline \end{array}$

Its magic constant is $216$.


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 Count Function of Power of Prime, $\map {\sigma_0} {p^k} = k + 1 \ge 9$.

Therefore we have $k \ge 8$.

$\Box$


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 $C$.

$\Box$


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 $C$.

$\Box$


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

$\Box$


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 $2$:

$C = p_1^{a_1} p_2^{a_2} k \ge 2^{a_1} 3^{a_2} \ge 2^3 3^3 = 216$

$\blacksquare$


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