last updated 8/5/02
This solutions page copyright © 2001, 2002 Paul C. Pasles
1. No! If the first row and the Vshaped bent row are both
equal to the magic sum, then the entries are not distinct. Proof by picture:

= 

, so 

= 

. 
In fact, it follows that there are (at most) only three distinct
entries. Can you see why?
2. The two possible answers are these:

and 

. 
3. YES. Here is just one example.
12  2  13  7 
15  5  10  4 
1  11  8  14 
6  16  3  9 
(There are many others.)
4. Good gravy! The magazine contains a slight typo, easily
fixed. The "A A^{T} " should read "A^{
T} A ". (In the other order,
you'd only get a semi magic product.)
Here are two ways of proving this odd little result.
First, the hard way. Clearly each symmetric pair sums to half
the magic sum. Say the magic sum is S = 2s. You can use the
various magic properties to fill every cell in terms of s and five
other variables, so that A has the form
a  b  s  b  s  a 
c  d  s  d  s  c 
e  c + e  d  s  c  e + d  s  e 
S  a  c  e  S  b  c  e  b + c + e s  a + c + e  s 
and you can multiply A^{T} by A using technology to get the desired result.
In case you don't feel like using the computer, here's another approach. Let's use vector notation. Denote the i ^{th} column of A by A_{i}, and write v.w for the dot product of the column vectors v and w. The i ^{th} column sum of A^{T}A is:
A_{1}.A_{i} + A_{2}.A_{i} + A_{3}.A_{i} + A_{4} .A_{i} = (A_{1}+A_{2} +A_{3} +A_{4}).A_{i} = [ 2s 2s 2s 2s ].A_{i} = 2s [ 1 1 1 1].A_{i} = (2s)(2s) = 4s^{2} = S^{2}.
The Vshaped and ^shaped bent rows work out similarly, using the property of vertical symmetry:
A_{1}.A_{1}+A_{2}.A_{2}+A_{2}.A_{3}+A_{1}.A_{4}
= A_{ 1}.(A_{1}+A_{4})_{ }+ A_{2}
.(A_{2}+A_{3} ) = A_{1}.[ s s s
s ]_{ } + A_{2}.[ s s s s
]
= (A_{1}+ A_{2}).[ s s s
s
] = [ s s s s
].[ s s
s s
] = 4s^{2} = S^{2}.
and
A_{4}.A_{1} + A_{3}.A_{2} +A_{ 3}.A_{3}+A_{4}.A_{4} = A_{4}.(A_{1}+A_{4})+ A_{ 3}.(A_{2}+A_{3} ) = ... = S^{2} .
The rows and the remaining bent rows can be worked out in a similar fashion, but there's a quicker way. Just recall from linear algebra that A^{ T}A is symmetric! (That is, symmetric in the usual linear algebra sense, that reflection along the main diagonal does not alter the matrix.) Then the rows and the remaining bent rows are automatically magic.
What about vertical symmetry? (That's symmetry with respect to the vertical bisector.) Check the sum of the i , j and i, n + 1j entries of A^{ T}A:
c_{i,j }+ c_{i,n+}_{1j} = A_{i}.A_{j }+A_{i}.A_{i,n+1j} = A_{i}.(A_{j }+A_{n+1j }) = A_{i}. [ s s s s ] = s A_{i}.[ 1 1 1 1] = (s)(2s) = 2s^{2 }= S^{2}/2,
which is half the magic sum of A^{T}A. That shows
the vertical symmetry!