There is no normal magic square of order 2

No $2 \times 2$ array can have the four distinct entries $1,2,3,4$ and be magic.

Suppose $M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ has all rows, columns and both diagonals summing to $s$. Comparing the first row $a+b=s$ with the main diagonal $a+d=s$ gives $b=d$, contradicting the requirement that the four entries be distinct. Hence no normal magic square of order $2$ exists.

This is the $n=2$ instance of the general fact that normal magic squares exist for every order $n \ge 1$ except $n = 2$.

Formalization Note IsNormal supplies injectivity of the index-to-entry map, and the two cells $(0,1)$ and $(1,1)$ are distinct, so the equality $b=d$ is immediately contradictory.