How to show there is no magic cube of order 2?

Question

I am reading Richard A. Brualdi's Introductory Combinatorics (Ch1.2).

Three-dimensional analogs of magic squares have been considered. A magic cube of order n is an n-by-n-by-n cubical array constructed out of the integers 1,2, ... , n3 in such a way that the sum s of the integers in the n cells of each of the following straight lines is the same: (1) lines parallel to an edge of the cube; (2) the two diagonals of each plane cross section; (3) the four space diagonals. The number s is called the magic sum of the magic cube and has the value (n4 +n)/2. We leave it as an easy exercise to show that there is no magic cube of order 2

I am not sure how to begin and would appreciate a tip on how to approach this

Thanks

Answer 1

Each layer of a magic cube must be a magic square. There is no magic square of order $2$, even with the numbers not in series. Write the square as $$ \begin {array}{c c}a&b\\c&d \end {array}$$ and note that $a+b=a+d, b=d$, similarly for other pairs, and all the numbers must be the same.

Answer 2

Hint: The magic sum will be $\frac{2^4 + 2}{2} = 9$, which means that each pair needs to contain exactly one odd number.