Some Theory concerning 5 x 5 magic squares
© H.B. Meyer

Consider a 5 x 5 magic square
c01
c02
c03
c04
c05
c06
c07
c08
c09
c10
c11
c12
c13
c14
c15
c16
c17
c18
c19
c20
c21
c22
c23
c24
c25

with 1 ≤ c01,c02,...,c24,c25 ≤ 25, all entries different, and equal row-, column- and diagonal sums (=65).

If all entries c01,c02,...,c24,c25 are replaced by 26-c01,26-c02,...,26-c24,26-c25 then the square remains magic.

Theorem 1   The entries of the magic square satisfy the following equations:

c05 = 65-c01-c02-c03-c04,
c10 = 65-c06-c07-c08-c09,
c15 = 65-c11-c12-c13-c14,
c17 = 2c01+c02+c03+c04+c06-c09+c11-c13+c16-65,
c18 = 325-4c01-2c02-2c03-2c04-2c06-2c07-c08-2c11-c12-c13-c14-2c16-2c19,
c20 = 2c01+c02+c03+c04+c06+2c07+c08+c09+c11+c12+2c13+c14+c19-195,
c21 = 65-c01-c06-c11-c16,
c22 = 130-2c01-2c02-c03-c04-c06-c07+c09-c11-c12+c13-c16,
c23 = 4c01+2c02+c03+2c04+2c06+2c07+2c11+c12+c14+2c16+2c19-260,
c24 = 65-c04-c09-c14-c19, and
c25 = 65-c01-c07-c13-c19.

 This results from solving the 12 linear equations rowsum=65, columnsum=65, diagonalsum=65 and expressing the 14-dimensional general solution with variables c01,c02,c03,c04,c06,c07,c08,c09,c11,c12,c13,c14,c16 and c19.

Remark 1   Every linear equation valid for the entries c01,c02,...,c24,c25 of all 5x5-magic squares, including the equations of Theorem 1, can be obtained with the "calculator" below (requires JavaScript):
c01 c02 c03 c04 c05
c06 c07 c08 c09 c10
c11 c12 c13 c14 c15
c16 c17 c18 c19 c20
c21 c22 c23 c24 c25



result:
Examples:
1) To get the first equation c01+c02+c03+c04+c05 = 65, simply press the "+"-button of row 1.
2) The equation 4c01+2c02+2c03+2c04+2c06+2c07+c08+2c11+c12+c13+c14+2c16+c18+2c19 = 325, f.i., can be obtained in the following way: three times "+" in row 1, two times "+" in rows 2, 3, and 4, one time "+" in row 5 and diagonal c01...c25, one "-" for columns 2, 3, 4 and diagonal c05...c21, finally two times "-" in column 5.
3) Another equation (*) 2c01+c02+c06+2c07-c14-c15-c18-c20-c23-c24 = 0, is shown by: "+" for diagonal c01...c25, column 1 and column 2, "-" for rows 3, 4 and 5.

Theorem 2   55 ≤ 3(c01 + c07) + 2(c02 + c06) ≤ 205.
This estimation is best possible, as shown by the square  
 1 2222515
 910161119
172313 5 7
2412 620 3
1418 8 421

 Proof  Equation (*) from example 3) gives c14+c15+c18+c20+c23+c24 = 2c01+c02+c06+2c07, therefore
55 = 1+2+3+...+10 ≤ c01+c02+c06+c07+c14+c15+c18+c20+c23+c24 = 3(c01+c07)+2(c02+c06). The upper bound is obtained, when every entry c is replaced by 26-c.

Corollary 1   20 ≤ c01+c02+c06+c07 ≤ 84.

 This follows from Theorem 2, since 58 ≤ 55 + c02+c06 ≤ 3(c01+c02+c06+c07). Then 60 ≤ 3(c01+c02+c06+c07), too. The square given in Theorem 3 shows, that the bounds 20 and 84 cannot be enlarged resp. reduced.

 Theorem 3    218 ≤ 3(c01+c07+c13) + 2(c02+c03+c05+c08+c11+c12) ≤ 328.
Proof   Define
n:= c01+c02+c03+c06+c07+c08+c11+c12+c13,   
s:= c04+c05+c09+c10+c14+c15,
f:= c19+c20+c24+c25.
n
s
 
f
 		     
22 6 31816
 414111521
 5 8122317
251319 7 1
 92420 210

One has n + s = 195 and f + s = 130, hence n = 65 + f ≥ 85 (see Corollary 1).
Now   n + c01+c07+c13 = 65 + c01+c07+c13 + f = 65 +c01+c07+c13+c19+c25+c20+c24 = 130 +c20+c24 ≥ 133.
Together with n ≥ 85 one now has 2n + c01+c07+c13 ≥ 133 + 85 = 218. The square above shows that the bound 218 ist best possible.

Theorem 4  For the "corner sum" the inequality 26 ≤ c01+c05+c21+c25 ≤ 78 holds.

Proof   From Theorem 1 (using the calculator of Remark 1) we get: 65 + c08+c12+c14+c18 = 2(c01+c05+c21+c25) + c13,    
hence 65 + 10 ≤ 2(c01+c05+c21+c25) + 25, and 50 ≤ 2(c01+c05+c21+c25). Equality is impossible,
since {c08, c12,c14,c18} = {1,2,3,4} would imply 26 = 5+6+7+8 ≤ c01+c05+c21+c25.

 Corollary 2  "X"-sum and "x"-sum: 52 ≤ c01+c05+2c13+c21+c25 ≤ 104.
 5 92025 6
1315 21124
21 123 317
1918 41410
 7221612 8
This follows, because exchange of proper rows and colums turns the "X"-sum c01+c05+2c13+c21+c25 into an "x"-sum c06+c08+2c13+c16+c18 and because "x"-sum + "corner sum" = 130. The example with "corner sum" 5+6+7+8 and "x"-sum 15+11+2·23+18+14 tells, that the bounds 26,78 in Theorem 4 and 52,104 in Corollary 2 cannot be improved.

 Theorem 5    c19 = 1/12(1495-19c01-9c02-7c03-10c04-9c06-11c07-3c08-2c09-9c11-5c12-5c13-6c14-8c16 ± sqrt(D)),
sqrt denoting the square root. D is the square number:
D =  -215c012-111c022-71c032-68c042-87c062-71c072-39c082-68c092-87c112-47c122-95c132-36c142-80c162
+c01(-258c02-190c03-172c04-210c06-134c07-30c08+124c09-210c11-98c12+70c13-12c14-200c16)
+c02(-138c03-132c04-126c06-90c07-18c08+84c09-126c11-78c12+66c13-12c14-120c16)
+c03(-100c04-90c06-62c07-30c08+52c09-50c12+22c13-12c14-90c11-80c16)
+c04(-84c06-44c07-12c08+40c09-84c11-44c12+52c13-24c14-80c16)
+c06(-90c07-42c08+36c09-126c11-54c12+42c13-12c14-120c16)
+c07(-54c08-4c09-66c11-58c12-34c13-12c14-40c16)
+c08(-36c09-18c11-18c12-42c13-12c14)
+c09(60c11+20c12-76c13-24c14+80c16)
+c11(-78c12+18c13-36c14-120c16)
+c12(-22c13-36c14-40c16)
+c13(-36c14+80c16)
+22750c01+15210c02+11830c03+10660c04+13650c06+10790c07
+5070c08-2860c09+13650c11+8450c12+650c13+3900c14+10400c16
-791375.

 This follows from the equation c012 + c022 + ... + c242 + c252 = 12 + 22 + ... + 242 + 252. Substitution of c05,c10,c15,c17,c18,c20,c21,c23,c24, and c25 by their expressions given in Theorem 1 leads to a quadratic equation for c19 with the above solutions.
Remark 2  The next two squares show, that the entries c01,c02,c03,c04,c06,c07,c08,c09,c11,c12,c13,c14, and c16 do not completely determine the magic square, but by Theorem 5 there are at most 2 squares having these entries in common, because there are not more than 2 possible values for c19. These values may be determined with the following "calculator" (JavaScript required, "NaN" means: "not a number"). 
20 11323 8
 42114 224
 6221216 9
101815 517
25 31119 7
 		   
20 11323 8
 42114 224
 6221216 9
101811 719
25 31517 5

c01 c02 c03 c04 c05
c06 c07 c08 c09 c10
c11 c12 c13 c14 c15
c16 c17 c18 c19 c20
c21 c22 c23 c24 c25


c19 =    


Relations for pattern sums    Consider the following sums of entries:

       
       
       
       
       
       
       
       
       
       
       
       
       
       
       
       
        
        
        
       
       
       
       
       
       
       
       
       
       
       
       
       
       
       
       
       
       
       
       
       
V=c01+c05
 +c21+c25		 v=c07+c09
 +c17+c19		 N=c01+c03+c05
 +c11+c13+c15
 +c21+c23+c25		 n=c07+c08+c09
 +c12+c13+c14
 +c17+c18+c19		 m=c13 K=c03+c11
 +c15+c23		 k=c08+c12
 +c14+c18		 O=c02+c04+c06
 +c10+c16+c20
 +c22+c24

Proper combinations of equations from Theorem 1 (with the help of the "equation-calculator" in Remark 1) lead to:

(R01) 2V+m = k+65 (R02) 2v+m = K+65 (R03) V+65 = n (R04) v+65 = N
(R05) 2v+3m+k = 195 (R06) 2V+3m+K = 195 (R07) 3V = v+2k (R08) 3v = V+2K
(R09) k = v mod 3 (R10) K = V mod 3 (R11) O = 3m+65 (In1) 26≤V,v≤78
(In2) 52≤V+2m,v+2m≤104 (In3) 91≤N,n≤143 (In5) 39≤k+3m,K+3m≤143 (In6) 44≤V+m,v+m≤86 *)

*) Inequality (In6) was found by a computer calculation.

These Relations are useful for showing, that certain entries in magic squares are impossible:

Example 1    There is no 5x5 magic square with c01=1, c05=2, c21=4, and c25=24.
 Proof    With (R01) the assumption leads to: m = k+65-2V = k + 3 ≥ 3+5+6+7 + 3 = 24. Because 24 is used: m=25. Further, by (R01) again: k=22 = 3+5+6+8. Now, c07+c19=15 leads to a contradiction, because there are not enough unused numbers left, that two of them could sum up to 15.

 Reduced 5x5 Magic Squares   The two maps
map1:
 		exchange rows 1 and 2, exchange columns 1 and 2
exchange rows 4 and 5, exchange columns 4 and 5
map2:exchange rows 2 and 4, exchange columns 2 and 4

in combination with the 4 reflections and the 3 rotations of a 5x5 square generate a set of 32 transformations (including the identity), each of which maps every 5x5 magic square onto another one.
Every 5x5 magic square can be mapped by one of these 32 transformations onto a unique "reduced square" for which the following nine inequalities are valid:
c01<c05, c01<c21, c01<c25, c01<c07, c01<c09, c01<c17, c01<c19, c05<c21, c02<c04.

The number of reduced 5x5 magic squares dependent on the value of c01 is shown by the table below:

c01number of reduced squares
117736163
215437677
312581407
48803796
55927144
63786042
72286273
81332837
9645631
10217929
1158598
1212809
 68826306

There exist 68820306 reduced 5x5 magic squares and 32x68826306 = 2202441792 magic squares of order 5.
(This number is well known since 1973, found by R. Schroeppel). There is a downloadable program "red5.exe" available, which produces all reduced 5x5 magic squares, moreover the user may prescribe fixed entries.

last update: 2010-October-24.






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