Herman Baer's Magic Square of Squares page

By Uri Raz

Report concerning magic square of squares (3x3), Jan/99

In "Scientific American" VIII.98 Martin Gardner in his quarter century review mentioned the above problem : rows and columns to have the same magic some of distinct squares as well as both the diagonals. Of the latter I wasnt sure in the beginning, but at any rateI prefered to content my self with the so-relaxed version of the problem.

On the other hand both solutions to the relaxed problem I succeeded to find have a square as a magic sum : there is foundation to suppose, not only by accident.

And here the solutions:

(1)
46 3 54 712
18 66 19 -"-
51 26 42 -"-
712 = = 5041
    (2)
34 27 66 792
18 74 21 -"-
69 6 38 -"-
792 = = 6241

each fulfilling 6 equations, as
(1') 462 + 32 + 542 = 712 = (46, 3,54) = (18,66,19) = (51,26,42)
      462 + 182 + 512 = 712 = (46,18,51) = ( 3,66,26) = (54,19,42)

(2') 342 + 272 + 662 = 792 = (34,27,66) = (18,74,21) = (69, 6,38)
      342 + 182 + 692 = 792 = (34,18,69) = (27,74, 6) = (66,21,38)

My first attempt for a general solution stated from the generalization of

(3)
2 9 4 15
7 5 3 -"-
6 1 8 -"-
15 = = 15

that is (a>b)

(3')
m-a m+(a+b) m-b S = 3m
m+(a-b) m m-(a-b) -"-
m+b m-(a+b) m+a -"-
S S S S S

and did not lead very far. Next I tried tostart from pythagorean triples, which led to some progress.

In the end I developed a procedure, based on the properties of numbers being represented as a sum of two squares, which allows to establish a practically unlimited collection of (3x3) arrays, for which the 3 rows and - say - the right hand column have the magic sum, while the two other columns dont, and differ from it by an amount D in opposite directions.

D can accept a lot of different values, among them 0; 1; 3; 5; 7; 8; 9; 11; 12; 13; 15; 16; ... So we get an interesting spectral distribution of D according to these values and their frequencies. so far for D up to 100:

1) not every possible value value is accepted, only 66, while 35 are missing.

2) multiples of 3 appear 26 out of 34 (average 7.2 cases), non multiples of 3 appear only 40 out of 67 (average 3.3 cases)

3) observing frequencies of accepted values of D generally, there are peaks at D=24 with 31 cases, at D=48 with 20, at D=96 with 17, and so on, while others appear just once, as D=12 or 13 for instance. (see diagram)

By the way square-numbers are represented throughout, with the exception of 4.

To describe the applied method more in detail, it is convenient to name the elements to be squared as follows :

(3a)
S c2 m2 c1=c c1^1 + m1^2 + c4^2 = S = c2^2 + m2^2 + c1^2 or B = S - c1^2 = c2^2 + m2^2 = c4^2 + m1^2 and so on. Let c4 > m1, c2 > m2.
S m3 m m1  
S c3 m4 c4  
  S S S S

So we choose a base B allowing at least two representations as a sum of two squares ; next we let c1 = C sum over successive values, receiving B + C2 = S as magic sum, which in the following is best demonstrated by a simple example : for instance if we choose

B = 52*13 = 325 = (182 + 12 = (18,1) = (17,6) = (15,1) ; then let us take C = 35 to get S = B + C2 = 325 + 1225 = 1550 as magic sum. So we have for the right hand columns three possibilities (35,1,18), (35,6,17), (35,10,15).

352 + 12 = 1226 = 2*613 has no other splitting into a pair of squares, 613 being prime, so no third row can be built in this case.

352 + 62 = 1261 = 13*97 has two representations and so allows an alternitive; but not so 352 + 172 = 1514 = 2*757, as 757 is prime. So we are left with (35,10,15) for the right hand column and with (18,7,35) or (17,6,35) for the upper row. Continuing on the same principles we arrive at a solution with D = 3:

52*13 = 325 B and 5 others with D = 11, 24, 24, 32, 59, in this case.
17 6 35 1550 S  
19 33 10 -"-  
29 22 15 -"-  
-3 +3 1550 22*73 M  
+/-D C^2 = k^2*M^2, k^2=2.37...

I started in then, and gradually, from different bases B beside

B1 = 52 * 13 = 325 [3], B2 = 52 * 17 = 425 [3], B3 = 5 * 132 = 845 [3],
B4 = 5 * 13 * 17 = 1105 [4], B5 = 5 * 172 = 1445 [3],
B6 = 53 * 13 = 1625 [4], B7 = 5 * 13 * 29 = 1845 [4],
B8 = 53 * 17 = 2125 [4], B9 = 52 * 132 = 4225 [4] = [5] - 1,
B10 = 52 * 13 * 17 = 5325 [6], B11 = 52 * 172 = 7225 [4] = [5] - 1,
B2a = 52 * 29 = 725 [3]

These I checked for rising values of C systematically. In [brackets] the number of splittings.

Introducing the mean-value M by S/3 = (B+C2)/3 = k*C2 we have

             k^2
 C^2 = B * -------, and I went up to some value k between √2 and √3
           k^2 - 3

Finally I checked as well several values of C for B3a = 52 * 37 = 925 [3] and B8a = 5^2 * 89 = 2225 [3] hoping (in vain) to get another solution with D = 0

Returnining to the two solutions with D = 0, S = 71^2 & S = 79^2, I gave them the following other arrangement:

(1'') 71; 66; 54; 51; 46; 42; 26; 19; 18; 3 S m (c1->c3) (c2->c4) (m4,m1) (m3,m1) \ / --- --- X <- -> (2'') 79; 74; 69; 66; 38; 34; 27; 21; 18; 6 S m (c3<-c1) (c4<-c2) (m2,m1) (m3,m4)

and correspondingly

(3'') 15; 9 8 7 6 5 4 3 2 1 S (m2 c4) (m3 c3) m (c1 m1) (c2 m4)

If we compare (1''), (2'') with (3'') and their structure in respect of the succession of cv and mv, they appear to be incompatible with one another, and possibly the "relaxed" solution could never be a solution to the unrelaxed problem (?).

As in the relaex problem all the nine elements are of equal status, we might have started from any element instead of giving preference to e, and we can put down an arrangement of equations like this:

 (1a') 71^2 - 66^2 =  5 * 137 =  685 =        5*137 [2] = (26, 3) = (19,18)
       71^2 - 54^2 = 17 * 125 = 2125 =       5^3*17 [4] = (46, 3) = (42,19)
       71^2 - 51^2 = 20 * 122 = 2440 =      2^3*5*6 [2] = (46,18) = (42,26)
       71^2 - 46^2 = 20 * 117 = 2925 =   3^2*5^2*13 [2] = (54, 3) = (51,18)
       71^2 - 42^2 = 29 * 113 = 3277 =       29*113 [2] = (54,19) = (51,26)
       71^2 - 26^2 = 45 *  97 = 4365 =     3^2*5*97 [2] = (66, 3) = (51,42)
       71^2 - 19^2 = 52 *  90 = 4680 = 2^3*3^2*5*13 [2] = (66,18) = (54,12)
       71^2 - 18^2 = 53 *  89 = 4717 =        53*89 [2] = (66,19) = (51,46)
       71^2 -  3^2 = 68 *  74 = 5032 =    2^3*17*37 [2] = (66,26) = (54,46)

 (2'a) 79^2 - 74^2 =  5 * 153 =  765 =   3^2*5*17 [2] = (27, 6) = (21,18)
       79^2 - 69^2 = 10 * 148 = 1480 =   2^3*5*37 [2] = (38, 6) = (34,18)
       79^2 - 66^2 = 13 * 145 = 1885 =    5*13*29 [4] = (38,21) = (34,27)
       79^2 - 38^2 = 41 * 117 = 4797 =  3^2*13*41 [2] = (69, 6) = (66,21)
       79^2 - 34^2 = 45 * 113 = 5085 =  3^2*5*113 [2] = (69,18) = (66,27)
       79^2 - 27^2 = 52 * 106 = 5512 =  2^3*13*53 [2] = (74, 6) = (66,34)
       79^2 - 21^2 = 58 * 100 = 5800 = 2^3*5^2*29 [3] = (74,18) = (66,38)
       79^2 - 18^2 = 61 *  97 = 5917 =    61 * 97 [2] = (74,21) = (69,34)
       79^2 -  6^2 = 73 *  85 = 6205 =    5*17*73 [4] = (74,27) = (69,38)

Here is set in evidence The algebraic advantage of S being a square. The underlined factorizations were actually used.

By the way if we tile the plane with one of these patterns every square of (3x3), which is cut out anywhere, has the (relaxed) magic property.


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