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.

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!
