Construct $4 \times 4$ magic square with fixed "1" 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.
 A: 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
A: 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!
