ABSTRACT Generalization of 4 4 Magic Square Deo Brat Ojha 1, KaulBL 1 Department of Mathematics, RKGIT, Ghaziabad Department of Mechanical Engineering, RKGIT, Ghaziabad deobratojha@rediffmailcom A magic square of order n is n n matrix containing integers in such a way that each row and column add up to the same value We generalize this notion to that of a 4 4 matrix with the help of a special geometrical figure without having much knowledge of algebra and another branch of mathematics Key words: Magic Square, Square matrix, Integer, Required sum 1 Introduction A magic square is an n n matrix filled with the integers in such a way that the sum of the numbers in each row, each column or diagonally also remain same, in which one integer use at once only Magic square and related classes of integer matrices have been studied extensively and still it is going on As we know, ancient India is the origin of numbers and its properties, then after it spreads all over the world The interested reader may consult the references therein[1,,3,4,5] But, the reader can easily extract that there is requirement of knowledge of algebra, number and it s properties and many different branches of mathematics for magic squares Here, we start with 4 4 matrix and with the help of special geometry able to obtain a magic square only having the knowledge of sum and multiplication of integers Preliminaries The sum of four square numbers in all directions can be achieved the same (equal) by using different figures, starting a number from one end to the other end clockwise or anticlockwise serially, we can achieve this easily by placing the figures in four different positions Any no 34,thirty four or above divisible by four(4) & remainder Two(), can be the sum of four squares in all directions By using 16,sixteen nos serially once only as below 16 4 5 9 34 15 3 10 6 34 14 7 11 34 1 13 1 8 34 34 34 34 34 34 34 9 6 11 8 34 5 10 7 1 34 4 3 14 13 34 16 15 1 34 34 34 34 34 34 34 For 16,sixteen squares a figure has been developed which can be placed on the squares & starting by the no got by dividing the whole sum no by four(4) & remainder two() & subtracted by seven(7) such as we want the sum of all the four different nos in all directions should be thirty four(34), divide 34 by 4 ; 34 4 = 8 & remainder = subtracting a constant 706
number seven(7) from eight( 8 ), we get one(1)we can start from 1 from one end of the figure clockwise or anticlockwise using 16 nos serially to get the sum of all different nos of four squares as 34 in all directions such as above 3 Main Result Example: A figure has been developed and if this figure is placed on 4*4 =16 square blocks clockwise or anticlockwise The some of all the lines in all directions can be achieved the same Figure given below Figure 1: Image showing figure developed with 16 square blocks If we want to get sum of the numbers as same in all directions, we must start with the number got by dividing it by four(4) and remainder two() up to sum 6 If we want a number 34 in all directions we must divide it with four(4) and the remainder should be two() ie (i) 34 4 = 8, remainder= by subtracting a constant number 7 from 8, 87=1, we can start with 1 serially on the figure & we will get thirty four,34 in all directions ie 707
16 15 1 4 3 14 13 5 10 7 1 9 6 11 8 (ii) Let the sum required is equal to 94 94 4 = 3, remainder=, by subtracting a constant number 7 from 3,37=16, we can start with 16 serially on the figure & we will get ninety four,94 in all directions ie 34 16 15 1 34 4 3 14 13 34 5 10 7 1 34 9 6 11 8 34 34 34 34 34 34 3 6 1 4 7 5 0 8 9 18 19 16 17 31 3 6 1 4 94 7 5 0 94 8 9 18 19 94 16 17 31 94 94 94 94 94 94 94 For the block of sixteen (16) squares, the following method can be obtained: 1 If the nos used have odd progression ( difference)between them, then after dividing the required sum with four the remainder will be two in these cases & the lowest sum is thirty four & the highest is infinity Example 1 (using difference of odd nos) 34 4=8 remainder =, starting point 87=1, difference=1, 94 4=3 remainder =, starting point 3=1, difference=3 8 1 13 1 34 11 7 14 34 6 10 3 15 34 9 5 4 16 34 34 34 34 34 34 34 34 37 1 94 31 19 40 4 94 16 8 7 43 94 5 13 10 46 94 94 94 94 94 94 94 If the numbers used have equal progressive even difference between them, the sum is divisible by four without any remainder & the minimum sum in this case is sixty four 64 & highest is infinity Example 1 (using difference of odd nos) 64 4=16, starting point 1615=1, difference=, 96 4=4, starting point 415=9, difference= 708
15 3 5 1 64 1 13 7 3 64 11 19 5 9 64 17 9 7 31 64 64 64 64 64 64 64 3 31 33 9 96 9 1 35 11 96 19 7 13 37 96 5 17 15 39 96 96 96 96 96 96 96 The subtracting factor of constant 7 is changed to 15 but after 6, it will be for every thirty it will keep changing like SL NO 1 3 4 5 6 7 8 9 10 NUMBERS USED 1,,3,4 1,3,5,7 1,4,7,10 1,5,9,13 1,6,11,16 1,7,13,19 1,8,15, 1,9,17,5 1,10,19,8, 1,11,1,31 1 3 4 5 6 7 8 9 10 LOW EST SUM 34 64 94 14 154 184 14 44 74 4 NO AFTER DIVIDINGW ITH 4 34/4=8 64/4=16 94/4=3 14/4=31 154/4=38 184/4=46 14/4=53 44/4=61 74/4=68 4/4=76 REMI NDER SUBTRACTIN G FACTOR CONSTANTS 7 15 7, 15, 7,,37 15,,45 7,,37,5 15,,45,60 7,,37,5,67 15,,45,60,75 DIFFERE NCE BETWEE N LOWEST SUM For the block of (4 4 ) square the numbers, which are using odd difference between them will be divided by four (4) & the remainder will be two() & if they are using even difference between them they are divisible by four94) without any remainder as 6,64,66 & 68 6 4=15 remainder= 64 4=16 remainder=nil 157=8 starting 1615=1 starting With odd difference ONE(1) 15 19 0 8 6 18 14 1 9 6 13 17 10 6 16 1 11 3 6 6 6 6 6 6 6 With even difference TWO() 15 3 5 1 64 1 13 7 3 64 11 19 5 9 64 17 9 7 31 64 64 64 64 64 64 64 66 4=16 remainder= 68 4=17 remainder=_ 167=9 starting 1715= starting 709
With odd difference ONE(1) 16 0 1 9 66 19 15 10 66 14 18 11 3 66 17 13 1 4 66 66 66 66 66 66 66 With even difference TWO() 16 4 6 68 14 8 4 68 1 0 6 68 18 10 8 3 68 68 68 68 68 68 68 The table given below gives some example of starting numbers & difference kept between them It is kept corresponding to the subtraction of constants 7,15,, & it is 1,,3,4 & so on For 7 it is 1,15 it is, it is 3 & it is 4 & so on SNo Required Sum Sum after divisible by 4 Reminde r Starting number after deduction of constant Differenc e Total sum of the numbers used in (4 4) is (required number 4) n Sn = { a+ ( n 1) d } where astarting ( number ) = 1 d( difference between numbers)= 1 34 8 87=1 1 where a = 1, d = 1 34 4 =136 4 10 107=3 1 where a= 3, d = 1 3 58 14 147=7 1 where a= 7, d = 1 4 6 15 157=8 1 where a= 8, d = 1 5 64 16 1615=1 where a=1,d= =56 6 70 17 177=10 1 where a=10,d=1 =56 7 96 4 415=9 where a=9,d= =56 8 154 38 9 184 46 387=31 1 where a = 31, d = 1 38=16 3 where a = 16, d = 3 4615=31 where a = 31, d = 46=16 4 where a= 16, d = 4 4645=1 6 where a= 1, d = 6 710
The total sum of 16,sixteen numbers in any combination remains same & is four times the required sum So we can find out the starting numbers & the difference between them by using the equation The equation formed as 4a + d = sum required The number divisible by four(4) with remainder two() should be subtracted by seven(7) or addition of fifteen(15) ie 7,,37,5 & so on To get the starting numbers & the difference between numbers The difference for 7,,37 & 5 is odd 1,3,5,7 & so on The numbers divisible by 4 without remainder should be subtracted by 15 or multiples of 15 to get different combination of starting numbers & difference between them, is even,4,6,8 etc For 15,,45,60 or by using the formula 4a +d = sum Required Let the sum required is 14 is divisible by 4 without any remainder So, the difference can be,4,6,8 & so on (even numbers) No required = 14 1 4a + d = 14, let d=, 4a + = 14,4a + 60 = 14, then a = 16 4a + d = 14, let d=4,4a + 4 = 14,4a + 10 = 14, then a = 1 No required = 14,4a + d = 14, let d=, 4a + = 14,4a + 60 = 14, then a = 16 38 40 16 14 36 8 4 18 14 6 34 0 44 14 3 4 46 14 14 14 14 14 14 14 No required = 14, 4a + d = 14, let d=4, 4a + 4 = 14,4a + 10 = 14, then a = 1 9 45 49 1 14 41 5 53 5 14 1 37 9 57 14 33 17 13 61 14 14 14 14 14 14 14 The above two combination can have six more combinations by using the figure developed which can be rotated clockwise or anticlockwise Let the sum required is 50, which is divisible by 4 with remainder So the difference can be 1,3,5,7 & so on, all numbers are odd 6 66 67 55 50 65 61 68 56 50 60 64 57 69 50 63 59 58 70 50 711
No required = 50 let d= 1 50 50 50 50 50 50 4a + d = 50, let d= 1,a = 55 4a + d = 50,let d=3,4a + 3 = 50, a = 40 4a + d = 50,let d=5,4a + 5 = 50, a = 5 61 73 76 40 50 70 58 79 43 50 55 67 46 8 50 64 5 49 85 50 50 50 50 50 50 50 60 80 85 5 50 75 55 90 50 50 70 35 95 50 65 45 40 100 50 50 50 50 50 50 50 4a + d = 50,let d=7,4a + 7 = 50, a = 10 The above four combinations can have 1 more combinations possible 59 87 94 10 50 80 5 101 17 50 45 73 4 108 50 66 38 31 115 50 50 50 50 50 50 50 4 Our Process Steps: IFix the required total sum(34 S < ), Then there exist two favourable cases (i) S = 4 p, p I (ii) S = 4 p +, p I II In case (i) Now, We have to decide starting number Number of blocks n1 = p [ w = { 1}] 71
n = n + d III Then calculate the sixteen numbers 1, where d is predefined by problem, if not then take d = 1 n, 1, n, n3,,, n 16 Later on w may be change after the fixed limit, but it will change in the manner W = nw + 1, n I IV Then arrange these sixteen numbers with the help of suggested geometrical figure V We can find numerous solution with the help of rotation of suggested figure in clockwise and anticlockwise direction VI We can find numerous solution by defining the suitable d VII But in all cases, we find optimized sum required VIII In case (ii), only d will be even with the same process IX Required figure 5 Properties of our Process I We may obtain various starting number for fix required sum II We may obtain various starting number with different difference for fixed required sum III IV We may get difference between numbers variable We may get different matrix with same set of numbers, with same difference via rotating the direction of figure from starting to end point in clock wise or anti clock wise direction and also can change the starting and end point for getting different matrix V So, we can start for finding magic square either with required sum or starting point or with difference or with rotation of figure Finally, we are working on 5 5 magic square matrix on the same line Further we will explore some more properties of this geometric figure in our subsequent article 51 Discussion on some applications The concept of balancing the magical squares with the help of suggested geometrical figure is easy and perfect This is a new concept for magical squares It is being used for moving bodies, Rotors, Cam shaft, Crank shaft etc 1 Cam shaft : Cam shaft having eccentric cam s at different angles and at different distances Cam s eccentric (out of centre) but are balanced with equidistance and angles 713
Crank shaft: The main journals of the Crank shaft are in the line of centre, but the crankpins are out of centre at different angles and distances It is balanced only by geometrical concept It is matter of further investigation towards it s possibility of usefulness of this concept in other frameworks 6 References 1 Harold M Stark An introduction to number theory MIT Press, Cambridge, Mass, 1978 Joseph H SilvermanThe arithmetic of elliptic curves SpringerVerlag, New YorkBerlin, 1986 3 Ezra, (001), BrownMagic squares, finite planes, and points of inflection on elliptic curves, College Math Journal, 3(4):60 67 4 Agnew, Elizabeth H, (1971) Two problems on magic squares, Mathematics Magazine, 44,1 15 5 Hanson, Klaus D, (1979) The magic square in Albrecht D urer s Melencolia I :Metaphysical symbol or mathematical pastime, Renaissance and Modern Studies, 3, 5 4 714