Numbers $1,2,\ldots,16$ are written in a $4 \times 4$ square matrix so that the sum of the numbers in every row, every column, and every diagonal is the same and furthermore that the numbers $1$ and $16$ lie in opposite corners. Prove that the sum of any two numbers symmetric with respect to the center of the square equals $17$.
Attempt:
Such a square matrix is known as a magic square. The magic constant is given by $M_n = \dfrac{n(n^2+1)}{2}$ and thus $M_4 = \dfrac{4(17)}{2} = 34$. Now assume without loss of generality that our square matrix is
$$\left(\begin{array}{cccc} 1 & a & b & c \\ d & e & f & g \\ h & i & j & k \\ l & m & n & 16 \end{array}\right).$$ We see that $e+j = 17$.
How do we prove the statement?
Suppose $c+l=17+q$. Then$f+i=17-q$ and (middle two columns) $a+b+m+n = 34+q$. However from top and bottom row, $a+b+c+l+m+n = 51 = 34+q+17+q \implies q=0$.
So this gives us $c+l = 17$ and $f+i=17$. That gives the diagonals proved.
I had less luck with the side elements. These may need a numerical "squeeze" due to the restricted number of possible integers... below are some thoughts as I jotted them down:
Now set $r = e+f \implies i+j = 34-r$ then
$\\ d+r+ g=34 \\k+(34-r)+h = 34 \implies k+h=r \\d+g+h+k =34 \\c+g+k=18 \\ d+h+(17-c) = 33 $
...