Problem source

A $3\times3$ magic square is a $3\times3$ array 
\[
\begin{array}{ccc}
a & b & c\\
d & e & f\\
g & h & k
\end{array}
\]
whose entries are the nine distinct integers $1,2,3,4,5,6,7,8,9$ and which has the property that all its rows, columns and main diagonals add up to the same number $n$. (Thus $a+b+c=d+e+f=g+h+k=a+d+g=b+e+h=c+f+k=a+e+k=c+e+g=n.)$
\begin{questionparts}
\item Show that $n=15.$
\item Show that $e=5.$
\item Show that one of $b,d,h$ or $f$ must have value $9$. 
\item Find all $3\times3$ magic squares with $b=9.$
\item How many different $3\times3$ magic squares are there? Why?
\end{questionparts}
{[}Two magic squares are different if they have different entries in any place of the array.{]}

Solution source

\begin{questionparts}
\item $(a+b+c)+(d+e+f)+(g+h+k) = 3n = 1 + 2 + \cdots + 9 = 45 \Rightarrow n = 15$.

\item Summing all rows, columns, diagonals through $e$ we have $(a+e+k)+(b+e+h)+(c+e+g)+(d+e+f) = 45 + 3e = 60 \Rightarrow e = 5$.
\item Suppose that one of the corners is $9$, then we need to find $2$ ways to make $6$ not using $5$ and $1$ (as $5$ is in the middle and $1$ diagonally opposite). Clearly this is not possible as the only remaining numbers are $2,3,4$ and only $2+4 = 6$. Therefore $9$ cannot be in the corner or central squares, ie it's one of $b,d,h,f$
\item  We must have
\begin{array}{ccc}
a & 9 & c\\
d & 5 & f\\
g & 1 & k
\end{array} and so $a$ or $c = 4$. Once we place $4$ by symmetry there will be another solution with $a = 2$. So:

\begin{array}{ccc}
4 & 9 & 2\\
d & 5 & f\\
g & 1 & k
\end{array} 

we now see $k$, then $f$, then $d$ then $g$ must be determined, ie:

\begin{array}{ccc}
4 & 9 & 2\\
3 & 5 & 7\\
8 & 1 & 6
\end{array} 
so our two solutions must be this and 
\begin{array}{ccc}
2 & 9 & 4\\
7 & 5 & 3\\
6 & 1 & 8
\end{array} 

\item For each of the $4$ possible placements of $9$ there are two magic squares, so there are $8$ possible magic squares, all related by reflection and rotation.
\end{questionparts}