On the existence of magic squares of every order different from $2$

Question

I was reading about magic squares and suppose that we speak here only of the magic squares that have in itself numbers from $1$ to $n^2$.

It is easy to see that we cannot have $2$x$2$ magic square because we cannot arrange numbers $1,2,3,4$ in such a square so that the sum of numbers in every row, column and diagonals is the same number.

But the natural questions that comes is:

Is it true (if it is proven, can someone point me to some references?) that for every $n \in \mathbb N \setminus \{2\}$ there exists at least one $n$x$n$ magic square?

Answer 1

Sierpinski in his classical book, Elementary Number theory, explains :

It is proved that there exist magic squares for any n >=3 (cf. L.Bieberbach).

referring to an article (in german) by Bieberbach (1954).

Moreover, Sierpinski claims to present a proof due to Makowski but the proof seems to depend on a conjecture (the so-called Schinzel conjecture or conjecture H in his book, cf. page 133).

Answer 2

Wikipedia gives a technique for constructing examples. For odd order, the elements from $kn+1$ to $(k+1)n$ are written on a (broken) diagonal so that each row and column contain one of each residue $\bmod n$ and one of each block $[1,n], [n+1,2n], \dots [n^2-n+1,n^2]$. The ones that are a multiple of $4$ are constructed by reflecting some of the numbers through the center. Those of the form $4n+2$ are formed from a square of size $2n+1$