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The method I have found to generate $4\times 4$ magic squares gives me a result in which the number "1" is at of the corners of the square. How can we extend this to a method to generate a magic square, for a fixed location of number "$1$"?

The number "$1$" can be in $16$ different locations (cells). If we name the cells, from upper left corner: $1, 2, 3,4, 5, \ldots,$ and the number "$1$" is at the $i^{th}$ cell, then how we can fill the other cells to make a magic square?

You can see variations here.

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There are really only three different locations: a corner, a side next to a corner, and one step diagonally in from a corner. If we can put $1$ into each of these, we can put it in any cell by using rotations and reflections. We already have a corner. We can subtract every number from $16$ and keep the square magic, but that doesn't help because it puts $1$ in the lower right corner. Because of the arrangement, we can add $8$ to all the numbers below $8$ and subtract $8$ from all those above, giving $$\begin {array} {c|c|c|c} 9&7&6&12 \\ \hline 4&14&15&1 \\ \hline 16&2&3&13 \\ \hline 5&11&10&8 \end {array}$$ That gets next to the corner. We can also rotate the $2 \times 2$ blocks by $180^\circ$ to get $$\begin {array} {c|c|c|c} 6&12&9&7 \\ \hline 15&1&4&14 \\ \hline 3&13&16&2 \\ \hline 10&8&5&11 \end {array}$$ which gets one in from the corner

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  • $\begingroup$ Thanks. Can we have a general solution for any doubly even order? $\endgroup$ – Susan_Math123 Aug 14 '17 at 4:49
  • $\begingroup$ The example you cited has a lot of symmetry that not all magic squares have. I exploited that in both of my transformations. They will not work starting with a generic doubly even square. $\endgroup$ – Ross Millikan Aug 14 '17 at 5:06
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At first let's define a more powerful magic square, which we will call $\color{Red}{\text{super-magic square}}$. By a $\color{Red}{\text{super-magic square}}$ we mean a magic square such the the sum of any arbitrary row is equal to the sum of any arbitrary column is equal to sum of any arbitrary diagonal.



For example suppose the following: $$\begin {array} {|c|c|c|c|} \hline 1&14&7&12 \\ \hline 15&4&9&6 \\ \hline 10&5&16&3 \\ \hline 8&11&2&13\\ \hline \end {array}$$

here we have:

$\color{Blue}{\text{Columns}}$: $$\begin {array} {ccccccccc} 1 & + & 15 & + & 10 & + & 8 & = & 34 \\ 14 & + & 4 & + & 5 & + & 11 & = & 34 \\ 7 & + & 9 & + & 16 & + & 2 & = & 34 \\ 12 & + & 11 & + & 3 & + & 13 & = & 34 \\ \end {array}$$

$\color{Green}{\text{Rows}}$: $$\begin {array} {ccccccccc} 1 & + & 14 & + & 7 & + & 12 & = & 34 \\ 15 & + & 4 & + & 9 & + & 6 & = & 34 \\ 10 & + & 5 & + & 16 & + & 3 & = & 34 \\ 8 & + & 11 & + & 2 & + & 13 & = & 34 \\ \end {array}$$

$\color{Purple}{\text{Diagonals parallel to the main diagonal}}$: $$\begin {array} {ccccccccc} 1 & + & 4 & + & 16 & + & 13 & = & 34 \\ 14 & + & 9 & + & 3 & + & 8 & = & 34 \\ 7 & + & 6 & + & 10 & + & 11 & = & 34 \\ 12 & + & 15 & + & 5 & + & 2 & = & 34 \\ \end {array}$$

$\color{Pink}{\text{Diagonals which are not parallel to the main diagonal}}$: $$\begin {array} {ccccccccc} 12 & + & 9 & + & 5 & + & 8 & = & 34 \\ 7 & + & 4 & + & 10 & + & 13 & = & 34 \\ 14 & + & 15 & + & 3 & + & 2 & = & 34 \\ 1 & + & 6 & + & 16 & + & 11 & = & 34 \\ \end {array}$$




$$ %% %% 1 + 14 + 7 + 12 = 34 , %% \\ %% 15 + 4 + 9 + 6 = 34 , %% \\ %% 10 + 5 + 16 + 3 = 34 , %% \\ %% 8 + 11 + 2 + 13 = 34 , $$

We will prove that, $1$ could be everywhere in a $\color{Red}{\text{super-magic square}}$.

Remark(I): Consider that a $\color{Red}{\text{super-magic square}}$ ($\color{Brown}{\text{of any arbitrary order}}$) is given. Then if we $\color{Blue}{\text{replace any two arbitrary columns}}$, then the resulting square, is again a $\color{Red}{\text{super-magic square}}$.

Remark(II): Consider that a $\color{Red}{\text{super-magic square}}$ ($\color{Brown}{\text{of any arbitrary order}}$) is given. Then if we $\color{Green}{\text{replace any two arbitrary rows}}$, then the resulting square, is again a $\color{Red}{\text{super-magic square}}$.


Now by $\color{Blue}{\text{column operations (I)}}$ and by $\color{Green}{\text{row operations (II)}}$ , we are able to change "the cell containing 1" to "any desired cell, so we are done!

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  • $\begingroup$ Where are the proofs/constructions for I and II? $\endgroup$ – OrangeDog Aug 14 '17 at 10:58
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    $\begingroup$ @OrangeDog: This is only valid if we ignore diagonals. The Wikipedia article linked from the question explicitly says that the main diagonals must have the same sum as each row or column, so this answer is not valid... $\endgroup$ – user21820 Aug 14 '17 at 11:00
  • $\begingroup$ @user21820 , Yes you are right. I have modified my answer, now this is true! $\endgroup$ – Jungle Boy Aug 14 '17 at 12:26
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    $\begingroup$ Alright it's okay for the 4*4 magic square now. The name is "Pandiagonal magic square." Also, you proved a weaker, not stronger, result, because it does not apply to all magic squares in general. $\endgroup$ – user21820 Aug 14 '17 at 13:24
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    $\begingroup$ And you should correct all your weird typos like "powerfull" and "rcolumn" and "coloumns"... $\endgroup$ – user21820 Aug 14 '17 at 13:26

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