Magic Square of Squares
Impossibility Proofs

3x3 Normal Magic Square of Squares
4x4 Normal Magic Square of Squares
5x5 Normal Magic Square of Squares
6x6 Normal Magic Square of Squares


3x3 Normal Magic Square of Squares Impossibility Theorem
A 3x3 magic square of squares of distinct integers in the range 1-9 is impossible.

Proof
The magic sum of squares is 95 which is 3 (mod 4).
The magic sum is odd, thus a magic series must contain an odd number
of odd integers.  But the sum of squares of 1 odd integer and 2 even
integers is not 3 (mod 4), as required.  So a magic series must contain
3 odd integers.  But 3 rows would contain 9 odd integers
and there are only 5 odd integers in the range 1-9.



4x4 Normal Magic Square of Squares Impossibility Theorem
A 4x4 magic square of squares of distinct integers in the range 1-16 is impossible.

Proof
The magic sum of squares is 374 which is 2 (mod 3).
The square of 0 (mod 3) is 0 (mod 3) and
the square of 1 or 2 (mod 3) is 1 (mod 3).
The only way that 4 squares can sum to 2 (mod 3) is if
exactly 2 are of the form 1 or 2 (mod 3).
So a magic series must contain exactly 2 integers of the form 0 (mod 3).
But 4 rows would contain 8 integers of 0 (mod 3) and there are only
5 integers of 0 (mod 3) in the range 1-16.



5x5 Normal Magic Square of Squares Impossibility Theorem
A 5x5 magic square of squares of distinct integers in the range 1-25 is impossible.

Proof
The magic sum of squares is 1105 which is 1 (mod 4) and 1 (mod 8).
The magic sum is odd, thus a magic series must contain
an odd number of odd integers.  But the sum of squares of 5 odd integers
is not 1 (mod 8) and the sum of squares of 3 odd integers and 2 even integers
is not 1 (mod 4).  So this leaves 1 odd integer and 4 even integers
as the only combination for a magic series.  But 5 rows would contain
20 even integers and there are only 12 even integers in the range 1-25.



6x6 Normal Magic Square of Squares Impossibility Theorem
A 6x6 magic square of squares of distinct integers in the range 1-36 is impossible.

Proof
The magic sum of squares is 2701 which is 1 (mod 4) and 5 (mod 8).
Since the magic sum is odd, a magic series must contain an odd
number of odd integers.  But the sum of squares of 3 odd integers and
3 even integers is not 1 (mod 4), as required.
So a magic series must contain 1 or 5 odd integers.

Let A be the number of rows with 1 odd integer.
Let B be the number of rows with 5 odd integers.
Since there are 6 rows,
   A + B = 6.
Since there are 18 odd integers in the range 1-36,
   A + 5B = 18.
Subtracting the first equation from the second yields
   4B = 12 or B = 3.
The only solution is A = B = 3, so there must be
three rows of 1 odd integer and 5 even integers combined with
three rows of 5 odd integers and 1 even integer.

For the three rows with 5 odd integers and 1 even integer,
the even integer must be 0 (mod 4) to make the sum of squares 5 (mod 8).
This describes the following magic series.
   (0,1,1,1,1,3) (mod 4)
   (0,1,1,1,3,3) (mod 4)
   (0,1,1,3,3,3) (mod 4)
   (0,1,3,3,3,3) (mod 4)
   (0,3,3,3,3,3) (mod 4)

Note that these contain no integers of the form 2 (mod 4).

For the three rows with 5 even integers and 1 odd integer,
the odd integer must be 1 (mod 4) to make the sum of squares 1 (mod 4).
This describes the following magic series.
   (0,0,0,0,1,2) (mod 4)
   (0,0,0,1,2,2) (mod 4)
   (0,0,1,2,2,2) (mod 4)
   (0,1,2,2,2,2) (mod 4)
   (1,2,2,2,2,2) (mod 4)

Note that these contain no integers of the form 3 (mod 4).

The six columns of the magic square of squares must also consist of this same
magic series pattern, three columns with the 3-series and three columns
with the 2-series.
A column must have one entry in common with each row,
so a column having the magic series (0,3,3,3,3,3) must have
a 3 in common with five different rows.  But there are only three rows
having a 3.  So this magic series cannot be used for a column.

By similar reasoning, the following magic series cannot be used in any row
or column because they require more than three of a 2 or 3 in common.
   (0,3,3,3,3,3) (mod 4)
   (0,1,3,3,3,3) (mod 4)
   (1,2,2,2,2,2) (mod 4)
   (0,1,2,2,2,2) (mod 4)

This leaves three rows taken from
   (0,1,1,1,1,3) (mod 4)
   (0,1,1,1,3,3) (mod 4)
   (0,1,1,3,3,3) (mod 4)
and three rows taken from
   (0,0,0,0,1,2) (mod 4)
   (0,0,0,1,2,2) (mod 4)
   (0,0,1,2,2,2) (mod 4)
as the required magic series combination.

There are nine integers of the form 3 (mod 4) in the range 1-36
which must all be contained within three rows.
The only way to get all nine is to use (0,1,1,3,3,3) three times.

There are nine integers of the form 2 (mod 4) in the range 1-36,
which must all be contained within three rows.
The only way to get all nine is to use (0,0,1,2,2,2) three times.

Thus, the magic series combination for the six rows must be the following.
   (0,0,1,2,2,2) (mod 4)
   (0,0,1,2,2,2) (mod 4)
   (0,0,1,2,2,2) (mod 4)
   (0,1,1,3,3,3) (mod 4)
   (0,1,1,3,3,3) (mod 4)
   (0,1,1,3,3,3) (mod 4)

The six columns must also use this same combination of magic series.
In particular, (0,0,1,2,2,2) must be used in three columns.
The three 2's in one of these columns can only be in common with entries
in the three rows containing 2's, thus the two 0's must be in common with
entries in the three rows containing 3's.  Three of these columns must have
a total of six 0's in common with the rows containing 3's,
but there are only a total of three 0's in those rows.
Therefore, a simultaneous row/column arrangement doesn't exist.