Prime Magic Square/Examples/Order 3/Smallest
== Example of Order $3$ Prime Magic Square == This order $3$ prime magic square has the smallest elements:  :$\begin{array}{|c|c|c|} \hline 67 & 1 & 43 \\ \hline 13 & 37 & 61 \\ \hline 31 & 73 & 7 \\ \hline \end{array}$  == Proof == For the purpose of this magic square only, we consider $1$ as a prime. A simple parity argument can show that $2$ cannot be included in a prime magic square: If it is, the row containing $2$ sum to an even number, while a row not containing $2$ will sum to an odd number. {{Improve|I'm drawing a blank on how to present the following result clearly}} We aim to show that all elements of an order $3$ prime magic square has the same remainder when divided by $3$. There are two parts to this: === $3$ cannot be used === For simplicity, we denote the numbers in the cells by their remainders when divided by $3$. Note that $3$ is the only prime divisible by $3$. We define the off-diagonals as: :$\begin{array}{|c|c|c|} \hline * &  &  \\ \hline  & * &  \\ \hline  &  & * \\ \hline \end{array} \begin{array}{|c|c|c|} \hline  & * &  \\ \hline  &  & * \\ \hline * &  &  \\ \hline \end{array} \begin{array}{|c|c|c|} \hline  &  & * \\ \hline * &  &  \\ \hline  & * &  \\ \hline \end{array} \begin{array}{|c|c|c|} \hline  &  & * \\ \hline  & * &  \\ \hline * &  &  \\ \hline \end{array} \begin{array}{|c|c|c|} \hline  & * &  \\ \hline * &  &  \\ \hline  &  & * \\ \hline \end{array} \begin{array}{|c|c|c|} \hline * &  &  \\ \hline  &  & * \\ \hline  & * &  \\ \hline \end{array}$ We also observe that, by switching rows and columns, the numbers in each row and column remains unchanged, while the two diagonals become two off-diagonals sharing one cell. Therefore the position of the numbers do not matter in the most part. Suppose $3$ is used in the square.  there are only two cases: ==== Case $1$: The row containing $3$ has numbers with all remainders ==== We have: :$\begin{array}{|c|c|c|} \hline 0 & 1 & 2 \\ \hline  &  &  \\ \hline  &  &  \\ \hline \end{array}$ Hence the row sum is divisible by $3$. Since: :$1 + 1 \equiv 2 \pmod 3$ :$2 + 2 \equiv 1 \pmod 3$ :$1 + 2 \equiv 0 \pmod 3$ there is a unique way to fill in the columns: :$\begin{array}{|c|c|c|} \hline 0 & 1 & 2 \\ \hline 1 & 1 & 2 \\ \hline 2 & 1 & 2 \\ \hline \end{array}$ Note that the order of $1$ and $2$ in the leftmost column do not matter due to symmetry. The sums of rows $2$ and $3$ are not divisible by $3$. Hence this case cannot occur. {{qed|lemma}} ==== Case $2$: The row containing $3$ leave out numbers with some remainder ====  suppose $2$ is not used. Then: :$\begin{array}{|c|c|c|} \hline 0 & 1 & 1 \\ \hline  &  &  \\ \hline  &  &  \\ \hline \end{array}$ Filling in the columns: :$\begin{array}{|c|c|c|} \hline 0 & 1 & 1 \\ \hline 1 & 2 & 2 \\ \hline 1 & 2 & 2 \\ \hline \end{array}$ All off-diagonals sum to $1$, which is not $1 + 1 = 2$. Hence this case cannot occur. {{qed|lemma}} === Primes of remainder $1, 2$ cannot be mixed ===  suppose there are $2$ $1$'s and $1$ $2$. Then the row sum is not divisible by $3$. We have: :$\begin{array}{|c|c|c|} \hline 1 & 1 & 2 \\ \hline  &  &  \\ \hline  &  &  \\ \hline \end{array}$ Filling in the first and third columns: :$\begin{array}{|c|c|c|} \hline 1 & 1 & 2 \\ \hline 1 &  & 1 \\ \hline 2 &  & 1 \\ \hline \end{array}$ Finally, filling up the rows: :$\begin{array}{|c|c|c|} \hline 1 & 1 & 2 \\ \hline 1 & 2 & 1 \\ \hline 2 & 1 & 1 \\ \hline \end{array}$ There must be an off-diagonal with sum divisible by $3$. Hence this case cannot occur. {{qed|lemma}} Using this result, we divide the primes $\le 73$ into two groups: :Remainder of $1$: $\set {1, 7, 13, 19, 31, 37, 43, 61, 67, 73}$ :Remainder of $2$: $\set {5, 11, 17, 23, 29, 41, 47, 53, 59, 71}$ We only need to show these primes cannot form a smaller magic square. Consider: :$\begin{array}{|c|c|c|} \hline a & b & c \\ \hline d & e & f \\ \hline g & h & i \\ \hline \end{array}$ Let $C$ be the magic constant. Then: {{begin-eqn}} {{eqn | l = 4C       | r = \paren {a + e + i} + \paren {b + e + h} + \paren {c + e + g} + \paren {d + e + f}       | c = These are all lines passing through the center }} {{eqn | r = \paren {a + b + c + d + e + f + g + h + i} + 3 e       | c = Center counted $4$ times }} {{eqn | r = 3 C + 3 e }} {{end-eqn}} Hence $e = \dfrac C 3$, which is $\dfrac 1 9$ of the sum of all numbers in the square. We have: :$1 + 7 + 13 + 19 + 31 + 37 + 43 + 61 + 67 + 73 = 352$ :$5 + 11 + 17 + 23 + 29 + 41 + 47 + 53 + 59 + 71 = 356$ $352$ and $356$ have remainders $1$ and $5$ when divided by $9$. In the lists: :$1, 19, 37, 73$ have a remainder of $1$ when divided by $9$. :$5, 23, 41, 59$ have a remainder of $5$ when divided by $9$. Omitting each number gives the corresponding center square values: :$39, 37, 35, 31$ for the first list :$39, 37, 35, 33$ for the second list Only $31, 37$ of the first list are possible candidates. However: :$73 + 31 > 93$ Hence $31$ fail to produce a magic square. This leaves $37$, which possibility is demonstrated above.  == Also see == * Magic Constant of Smallest Prime Magic Square == Sources == * {{MathWorld|Prime Magic Square|PrimeMagicSquare}}