Fun with Mathematics

Numeral relations of kind 2R²₂ with numbers ending with 1

One can construct numeral relations of kind 2R²₂ with numbers ending with any number. The general form for ^2R22 with numbers ending with 1 is given by

(10a+1)²  + (10b+ 1)²  = (10c +1)² + (10d+1)²

It can be shown that it will exist under the condition

(a-c)[5(a+c) +1] = (d-b)[5(b+d) +1]

By giving any desired number ending with 1 to a and c, the corresponding amicable values for b and d can be determined with the help of the above condition. For example, when a=7 and c= 1,we have,

6 x41 = 246 = (d-b)[5(d+b)+1] .246  can be expressed as the product of two factors in different ways- 1x246 ,2x123,3x82. Only suitable pair of factors will give integral solutions for b and d.

With 1 x 246 we get d =25 and b = 24 so that  71² + 241² = 11² + 251²

With other pairs of factors,we arrive at numeral relations with numbers ending with different digit.

With 2 x 123  we have, 71²  + 113²  = 11²   + 133²

3 x 82 gives , 71² +  67²   = 11²   +97²

From these relations we see that when two factors of a product remain same, one square root number

In either side of the relation is unaltered.  By subtracting one from the other ,one can generate more and more  2R²₂ relations. With the help of the above relations we get,

113² + 97² = 133² + 67²

241²+ 133² = 113² + 251²

241² + 97² = 67² + 251²

By following similar procedure, one can develop 2R²₂ relations with all square root numbers ending with 1. When a = 17,27,37,…..(10n +7) and c = 1 we get such relations.

 If a= 17 and c= 1, then b= 145 and d= 146

               171² + 1451² = 11² + 1461²

If a = 27 and c = 1, then b = 366 and d = 367,

271² + 3661² = 11² + 3671²

When a= 37 and c= 1 then, b = 687 and d= 688,

371² + 6871² = 11² + 6881²

One can generalize this method with the factors of  a product. The equivalent products of two factors and the conditions for a specific requirement are given by

      f₁ x f_{2 =} f₃ x f₄

(a-c)[5(a+c)+1] = (d-b)[5(b+d)+1]

By solving we arrive,

10a = f₂ – 1 +5f₁ ; 10b = f₄-1 -5f₃  ; 10c= f₂ -1 -5f₁ ; 10d = f₄-1 +5f₃

By substituting these values in the primary relations assumed, we get,

(f₂ +5 f₁)² + (f₄ – 5f₃)² = (f₂ – 5f₁)² + (f₄ + 5f₃)²

It implies that if the two factors of a product end with digits 1 and 6 respectively , one can generate quite a large number of 2R²₂ relations with all numbers end with 1.

 If 6x31= 1x186 ; 61² + 181² = 1² + 191²

     6 x 51 = 1 x 306 ; 81² + 301² = 21² + 311²

   6 x 61 = 1 x 366 ; 91² + 361² = 31² + 371²

By  adopting this simple procedure, one can generate 2R²₂ relation with all numbers ending with any particular digit. The 2R²₂ relation with square root numbers ending with n is given by

(10a+n)²  + (10b + n)² = (10 c + n )² + (10d + n)²  and the required condition is

(a-c)[5(a+c)+n] = (d-b) [5(b+d)+n]

Fun with Mathematics

More about Magical properties of Magic square

Magic squares of any order can be constructed with a set of numbers, so that its sum in rows,

columns and main diagonals will be same.  They have many curious mathematical properties. 

Such a magic square may be assumed that its small squares are uniformly filled with weights

Instead of numbers, (The magnitude of weight is exactly equal to the corresponding number).

When this magic square is pivoted at its CG, it is found that it is in an unstable equilibrium.

Under this condition, net moment about any symmetrical axis will be equal to zero. That is,

the clock wise moment will be equal to the anti-clock wise moment.

This can be verified with a typical example, a 3 x 3 magic square filled with natural numbers from

1 to 9.

                                                                                                      x

2	3½	3½	6
9	2½	2½	1
4	1½	1½	8

                                                                                                      y

With respect to the vertical axis xy , both the clock wise and anti-clock wise moment are equal and

is equal to 7½  X + 15 Y, where

X and Y  are the perpendicular distance from the axis xy to the CG of the component weight.

The equidistant rows and columns with respect to symmetrical axis are interchangeable, where as

odd row or column, if any,

will be common, The interchangeable columns and rows are having not only equal sum but also the

sum of their of squares will also be equal.

2+7+6= 4+3+8 = 15  and  

 2² + 7² + 6² = 4² + 3² + 8² = 89

 Since the middle column is common, its value is equally shared by both the interchangeable

columns. Considering this equal shares,

We have,

5½ + 11½ + 5½ = 9½ + 3½ + 9½ = 22½

The sum of their squares again equals

 2(5½)²  + (11½)²  = 2 (9½)² + (3½)² = 192.75

For the interchangeable rows,

6½+ 9½ + 6½  = 8½ + 5½ + 8½

2(6½)² + (9½)²  = 2 (8½)² +  (5½)² =  174.75

For a main diagonal (ascending), by cancelling out the moment due to weights in the common

diagonal, we have,

7(d/2) + 2d + 9(d/2) 1(d/2) + 8d + 3(d/2) = 10 d

Even though they are interchangeable sets ,they are not equidistant, and hence the sum of their

squares will not be same.

For any system under equilibrium, the moments with respect to any pair of vertical or horizontal axes

passing through edges will be equal.

Since the perpendicular distances between the reference axis and the weights in a row/column are

x,2x,3x , we can derive

For vertical axis ,right edge; 2x+14x + 18x =34x

                                               9x+10x+3x = 22x              or 2(34x) +22x = 90 x

                                               4x+6x+24x = 34x       

For vertical axis,left edge; 6x+14x+6x= 26x

                                            x+10x+27 x= 38X               or 2(26x) + 38x= 90 x

                                           8x +6x+ 12 x = 26x      

As the set of numbers is interchangeable and equidistant, we have the additional square relation

2(34)² + (22)² = 2 (26)² + (38)²  = 2796

The horizontal top and bottom axes give,

2(32x) + 26x = 2(28x) + 34x = 90x

2(32)² + (26)² = 2(28)² + (34)² = 2724