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.
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== 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 $k \ge 8$.
=== 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 $2$:
:$C = p_1^{a_1} p_2^{a_2} k \ge 2^{a_1} 3^{a_2} \ge 2^3 3^3 = 216$
== Sources ==
* : $216$
* : $216$
Category:Multiplicative Magic Squares