Observations and Proofs:
Columns, General Observations
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COLUMNS, GENERAL OBSERVATIONS
- All columns are magical.
Without loss of
generality, notice from the construction algorithm that - in
moving rightward column-by-column across the square - at each
move, with the exception of the seed or end cells, values
increase and decrease, respectively, in the n/2 plus and
the n/2 minus rows. The values change in the same way when
moving into either of the seed cell columns, with the
exceptions noted above. Consequently, since each row has
both a seed and an end cell and there are an equal number
of plus and minus rows, n2 is added to what would otherwise
be the values of as many cells in each of the seed columns
as it is subtracted from. Thus, we can neglect the
influence of the seed cell columns adjustment in this
Reconsider now what happens to the values of
cells when moving, column-by- column, across a square.
First note that the value of the cell in a row will either
be increased or decreased by either the first or second
element of that row's root pair.
Without loss of
generality, we will assume that the value is changed by the
second element. This is 2x - 1, where x is the seed cell
value for the row. Observe from the construction algorithm
that - for exactly half of the rows - the (n/4 + 1)st. to the
(3n/4)th rows, inclusive, the seed cells increase in value
by one with each subsequent row. This means that the sum of
the x values in the minus rows exceeds that of theplus rows
in one half of the square by (n/2)/2 = n/4. From this, we
can infer that the net value of the cells in this half of
the square decreases by 2(n/4) = n/2 when moving rightward
from one column to the next, since (2z - 1) - [2(n/4 + z) -
1] = - 2(n/4) = - n/2, where z is the sum of the value of
the plus rows in the central half of the square.
The opposite situation prevails in the remaining half of the
square's rows. Thus, the net effect on the value of a
column's cells when moving across the square is nil.
Finally, since the value of the seed column pair must be
n(n2 + 1), (Observation 3 and the construction algorithm),
the value of each of its columns (as well as that of every
other column) must be (n/2)(n2 + 1). Thus, every column is
- The mean value of the cells in each column is
(n^2 + 1)/2.
This follows directly from Observation 11
and the fact that each column has n cells.
- Every order-2 subsquare (core) is (4/n)-magical.
If the core is located in the seed columns, then its value is
obviously 2(n2 + 1) = (4/n) *(n/2)(n2 + 1), or (n/4)th of the
magic constant, so assume that it is located elsewhere.
Without loss of generality, assume that its two rightmost
cells are an even positive number of cells, say x cells, from
their rows' left seed columns cells. Then the core's leftmost
cells are x - 2 cells from their rows' right seed cells. Hence,
recalling that the difference in value of every other cell is 2n,
with the exception that n2 is added to (in a minus row) or
subtracted from (in a plus row) what would otherwise be the
value of the right seed cell (Observation 2), we see that the
value of the core's lower row is:
n2 + 1 + (x/2)(2n) + ([x - 2]/2)(2n) - n2 (if it is in an addition
n2 + 1 - ((x/2)(2n) - ([x - 2]/2)(2n) + n2 (if it is in a subtraction
Since the core's two rows are adjacent, they must be of different
types. Consequently, if one of the above values is that of the core's
lower row, the other must be that of its upper row (and vice-versa).
A similar argument can be made if the core's leftmost cells are an
even positive distance from their rows' left seed cell. Hence, the
value of a core given the above conditions is the sum of (1) and (2),
or 2(n2 + 1), the same as that of a core located on the seed columns.
Consequently, a core is (4/n)-magical.
- If a square's order is an even power of 2, it contains magical
minisquares (but not half-magical ones); if its order is an odd
power of 2, it contains half-magical minisquares, (but not magical ones).
Recalling that the order of a Franklin square is 2p where p > 2, let
o be the order of a square and p the power of 2 it equals . If o is a
square number, then p is even, and minisquares of o, but not of o/2 cells,
exist. If o is not a square number, then p is odd, and minisquares of o/2,
but not of o cells, exist. Since any minisquare of n or n/2 cells, respectively,
can be decomposed into n/4 or (n/2)/4 = n/8 cores, each of which is (4/n)-magical,
any minisquare of n or n/2 cells, respectively, is magical or half- magical.
Thus, if a square's order is an even power of 2, it contains magical minisquares
(but not half-magical ones); if its order is an odd power of 2, it contains
half-magical minisquares (but not magical ones).
- (i.) Cell pairs in adjacent columns and alternating rows are equivalent.
(ii.) Cell pairs in adjacent rows and alternating columns are equivalent. (i.)
Imagine two overlapping cores, A and B, with A's lower row being B's upper.
Since all cores are equivalent and a number is equivalent to itself, we can
infer that the upper cell pair of A and lower cell pair of B are also equivalent. As this
can occur anywhere on the square, any pair of adjacent cells must be equivalent to
all same-column cell pairs in alternating rows. Thus, all cell pairs in adjacent
columns and alternating rows are equivalent.
(ii.) The same reasoning applies to alternating column same-adjacent-row cell
Thus, all adjacent rows cell pairs in alternating columns are equivalent.
Every congruent quadruple of cells is half magical.
On the square table below, consider the sixteen innermost cells
(Region 1). From Observation 14 we know that cell pairs 28,36
and 30,38 and cell pairs 27,35 and 29,37 are equivalent. Similarly,
cell pairs 20,21 and 36,37 and 28,29 and 44,45 are also equivalent.
Since cell pairs 28,36 and 29,37 form a core, we can infer from the
first set of equivalencies that cell pairs 30,38 and 27,35 are together
equivalent to one. Similarly, we can deduce that cell pairs 20,21 and
44,45 are also equivalent to a core. Thus Region 1 - which must,
from its size, be equivalent to four cores - has, less its corner cells,
a value equivalent to three. Thus, its corner cells are equivalent to a core.
Now add to this region the remaining lined cells. This new region is
equivalent to nine cores. The cells without horizontal lines in them
can be decomposed into groups equivalent to cores, as shown.
Imagine removing these and the clear cells from the grid and bringing
the remaining (horizontally lined) cells together. Look familiar?
Your observation and the preceding reasoning let you infer that the
corner cells of the non-clear region are also equivalent to a core.
We can generalize this result to any even-ordered square region.
Thus, every quadruple of cells is (n/4)-magical.