If sum of two numbers equals with sum of two other numbers , then their product will never be equal . If a+b = c+d ,then ab ≠cd The viceversa is also true.if ab = cd then a+b ≠c+d.

Let L and B be the length and breadth of a rectangle. Then its perimeter become 2(L+B) = P and its area A = LB.Another rectangle with same  P  has length L’ – (L-a)and breadth B’= (B+a). To have same area A’ = L’B’ =(L-a)(B+a) = LB +a(L-B) – a² = LB. Or (L-B) = a which demands  L = B+a and B = L-a where the length and breadth are interchanged.. 

P = 2L + 2B.  and area A = LB.When P remains same B = (P/2) – L, the ares of the equi-perimeter rectangles A = L[(P/2) – L] = {PL/2) – L^2. .The chnnge in area due to the change in the dimension of the rectangle having same perimeter dA/dL = (P/2) – 2L A is maximum when dA/dL = 0 or P = 4L which demands L =B  

Let L be the side of a square. Then its area A = L² and its perimeter P = 4L. When it is reshaped into a rectangle with length L’ = L+x and breadth B” = L-x. Then the area of the rectangle A’ = L’B’ =(L+x)(L-x)= L² – x² = L² or x = 0..The area of a rectarangle will be maximum when it becomes a square with same perimeter.

 Improve your mathematical skill

(1) {3,4=5]² is the smallest Pythagoras triple. What is the next higher Pythagoras triple having same ending digits ?

 (2) Find a solution for three successive odd numbers whose sum gives a cube?

 (3) 9 +9 = 18  ; 9 x9 =  81 Here the sum and the product of two numbers give results where the digits are revered. Can you find out more examples?

(4) x⁵ + x⁴ + x³ + x² + x + 1= 0. Find the value of x.

 (5) xyz + xyz + xyz = zzz or xyz + xyz + xyz = yyy or xyz + xyz + xyz = www, where w,x,y,z take any digit from 1 to 9. Find allowed values of w,x,y.z.

 (6) (3+2) / (3+1) = (3³ + 2³) / (3³ + 1³) and (6+4) / (6+2) = (6³ + 4³) / (6³ + 2³)  Find an another example.

(7) Under what condition the relation (a+b)/(c+d) = (a² – b²)/(c² – d²) is true. Give numerical example

 (8) Prove that sum and product of two numbers cannot be equal to sum and product of two other numbers simultaneously.

(9) If x² + 1/x² = 34 . Find the value of x⁵  + 1/x⁵

 (10) Find a set of Pythagoras triples with given endings (8.5 and 7)

 Answers

(1) [13,84=85]²

        13 is prime;13² = (c-b)(c+b) ,if c-b=1 and c+b = 169 ,  c = 85 and b = 84  

 (2) (2n-1) + (2n+1) + (2n+3) = 6n +3 = 3 (2n+1). To give odd cube (2n+1) must be odd  or (2n+1) = 9, 243 ------9 (2x-1)³  . It gives

7+9+11=27 = 3x3x3 = 3³  

241 + 243 + 245 = 729 = 9x9x9 = 9³

(3) Let m and n be the two numbers satisfying the required condition. m+ n = 10 x + y and mn = 10y + x where x, y are any single digit number.                                                                                        m = 10x + y – n  = (10y +x )/n                                                                                                            which gives n²  = x(10n -1)- y(10-n)                                                                                                       To make n to be positive  x(10n-1)  > y(10-n)                                                                                        n=1;  1 = 9(x-y) , there is now whole number solution                                                                                         n = 2; 4 =  19x – 8y, which gives  x =4 and y =9  ,or m = 47                                                                         2 + 47 = 49;  2 x 47 = 94                                                                                                                n=3 ; 9 = 29x – 7y, which provides x =2. Y =7 or m = 24                                                                                3 + 24 = 27  ; 3 x 24 = 72                                                                                                                    For higher values of n , n = 4,5,6… ., there is no such pair of numbers .                                    If the sum and product are three digit numbers, then  m+n = 100x + 10 y + z and mn = 100z + 10y +x                                                                                                                                                                   m = (100z +10y +x)/n = 110x + 10y +z – n                                                                                                            n² = x(100n -1) +10y (n-1) – z(100-n)                                                                                                  when n = 2 ; 4 = 199x + 10y – 98z                                                                                                           when x = 4 , z= 9  we have 786 + 19y – 882  which gives y = 9 0r m – 497                                   2 + 497 = 499 ; 2 x 497 = 0=994  This is true for any number of 9 in between 4 and 7.                              2 + 4997 = 4999 ; 2 x 4997 = 9994                                                                                                           2+49997 = 40000 ; 2 x 49997 = 99994

(4) The given expression can be written as x (x⁴ + x³ + x² + x + 1) = -1. There are two possibilities.

x = 1 and (x⁴ + x³ + x² + x + 1) = -1   ;   x = -1 and (x⁴ + x³ + x² + x + 1) = 1

The former solution is not valid as (x⁴ + x³ + x² + x + 1) ≠ -1 if x =1. But x = -1, satisfies both the conditions. The solution is same to all problems where the exponents of x begin from 0  to (2n-1)

x³ + x² + x + 1 = 0                                                                                                                                       x⁷ + x⁶ + x⁵ + x⁴ + x³ + x² + x +1 =0

(5) xyz + xyz + xyz = zzz

3 (100 x + 10 y+ z) = 100 z + 10 z + z = 111 z

      100x +10y + z = 37 z -à  100x +10y = 36 z à 50 x + 5 y = 18 z

The digital roots must be same in both sides. Whatever may be the digital root of z, because of its coefficient 36 , 36z will have a digital root 9. Hence the digital root of x +y must be 9.

One possible solution is z =5 , x= 1 and y =8)185, which gives 185 x 3 = 555

xyz + xyz + xyz = yyy

3 ( 100 x + 10 y+ z) = 100 y + 10 y + y = 111 y

     100 x + 10y + z = 37 y --à 100x + z = 27 y

The digital roots must be same in both sides. Whatever may be the digital root of y, because of its coefficient 27,  27 y will have a digital root 9. Hence the digital root of x +z must be 9

One possible solution is  y =4 ,x=1 , z= 8 which gives 148 x 3 = 444

( x≠w, as it gives forbidden solutions)

If the identical digit (w) in the sum is different from the digits of the adder (x,y,z)

3(100x + 10y +z) = 100 w + 10w + w = 111 w

(100x + 10y +z) = 37 w

37 x 3n gives all identical digits. Two solutions are available\

259 x 3 = 777 and 296 x 3 = 888

 (6) We know a³ + b³ = (a+b)(a² – ab + b² ) and a³ + c³ = (a+c)(a² – ac + c² )

If a² – ab + b² and a² – ac + c² are equal  (a+b)/ (a+c) =( a³ + b³ )/ (a³ + c³ )

The condition demands a² – ab + b² = a² – ac + c²

 a(c-b) = c² – b² = (c-b) (c+b)  or a = c+b or c = a—bThe general form of this relation is

(a +b)/ [a +(a-b)] = (a³ + b³ )/ a³ + (a-b)³ . When a = 7 , b =5, we have  (7+5)/ (7+ 2) = (7³+5³)/ (7³+ 2³)

(7) (a+b)/(c+d) = (a² – b²)/(c² – d²) = (a+b)(a-b) / (c+d)(c-d) .To make them to be equal a-b = c-d is the required condition. For example, a=7,b=5 , a-b = 2, c = 3 , d = 1 and c-d =2, we have (7 +5)/(3+1) =  12/4 = (7² – 5²) / (3²  - 1² )  = 24/8

(8) Let us suppose  a + b = c+d and ab = cd. Squaring the former relation a² + b² + 2ab = c² + d² + 2cd and (a + b)³ = (c+d)³ -à a³ + b³ + 3ab ( a+b) = c³ +d³ + 3cd (c+d), Since ab = cd,  a² + b² = c² + d² and a³ + b³ = c³ + d³

If a + b = c + d, the difference between one pair (a-c) must be equal to the difference in the other pair (d-b) That is if  c = a +x , then d = b – x. Substitute these values in the relation ab = cd = (a+x)(b-x) = ab + x(b-a) – x²  which gives a condition x = b-a. Substituting this value in c = a + x = a + b – a = b and d = b – x = b – b + a or d = b. The condition demands that two numbers must be same in both sides of the relation. It proves that if sum of two numbers equals to sum of two other numbers, then their product will not be equal, or if the product of two numbers equals to product of two other numbers, then their sum will not be equal.

(9) (x + 1/x)² = x² + 1/x² + 2 = 34 + 2 = 36                                                                                             Or x + 1/x = ± 6. With positive root, it gives a quadratic equation x² – 6c + 1 = 0. Its solution is x = 3 ±2√ 2                                                                                                                                                         (x + 1/x)³ = x³ + 1/x³ + 3(x + 1/x)                                                                                                           6³ = 216 = x³ + 1/x³  + 18 or  x³ + 1/x³  = 216 – 18 = 198                                                                                      (x² + 1/x²)² = (34)² = 1156 = x⁴ + 1/x⁴ + 2 which gives x⁴ + 1/x⁴  = 1154                                                  (x⁴ + 1/x⁴)(x + 1/x) = 1154 x 6 =6924 = (x⁵ + 1/x⁵)  + (x³ + 1/x³) which gives                                               (x⁵ + 1/x⁵) = 6924 -   (x³ + 1/x³) = 6924 – 198 = 6726

This can also be derived from (x + 1/x)⁵                                                                                                                                  (x + 1/x)⁵  =6⁵ = 7776 = (x⁵ + 1/x⁵) + 5 (x³ + 1/x³) + 10 (x + 1/x)                                                      (x⁵ + 1/x⁵) = 7776 – 5(198) – 10(6) =6726                                                                                   Extending this idea (x⁶ + 1/x⁶)  = (x³ + 1/x³)² – 2 = 198² -2 = 39202

(10) The general form of the Pythagoras triple is (10n +8)² + (10x +15)² = (10x + 17)² which gives a condition 5n² + 8n = 2x.For all even n ,x becomes a positive number. When n = 2 , x = 18, which gives [28,195 = 197]² and when n = 4 , x = 56 which gives [48,575=577]²

The general form of this set is (10n +8)² + (25n² +40n +15)² = (25n² +40n + 17)²

problem solving

 As a mixed power relation show that  (x)⁵ + (x)⁵ = 2(x)⁵ can be expressed  as a power with exponent other thn 5 or multiples of 5

Let x = 2ⁿ k^{m ,} then  (2ⁿ k^m)⁵ + (2ⁿ k^m)⁵ = 2(2ⁿ k^m)⁵ = 2⁵ⁿ⁺¹ k^5m

when n = 1 ,m=2 ; (2 k²)⁵ + (2 k²)⁵ = 2(2 k²)⁵ = (2³ k⁵)²

Keeping n same and m is changed

when n = 1 ,m=4 ; (2 k⁴)⁵ + (2 k⁴)⁵ = 2(2 k⁴)⁵ = (2³ k¹⁰)²

Keeping m is same and n is changed

when n = 3 ,m=2 ; (2³ k²)⁵ + (2³ k²)⁵ = 2(2³ k²)⁵ = (2⁸ k⁵)²

when n=1 m=3 ; (2 k³)⁵ + (2 k³)⁵ = 2(2 k³)⁵ = (2² k⁵)³

when n=3 m=4 ; (2³ k⁴)⁵ + (2³ k⁴)⁵ = 2(2³ k⁴)⁵ = (2⁴ k⁵)⁴

when n=1 m=6 ; (2 k⁶)⁵ + (2 k⁶)⁵ = 2(2 k⁶)⁵ = (2 k⁵)⁶

In 2⁵ⁿ⁺¹ k^5m the exponents 5n+1 cannot be expressed as a multiple of 5. Hence sum of two power number with exponent 5 cannot be expressed as a number with same exponent. This is in accordance with FLT.

 

.Find the sum  of series of even squares in natural series .

S²_e = 2² + 4² + 6² + 8² ……… (2n)²

        = 2² [ 1² + 2² + 3² ………. n² )   where n = N_H/2

     = 4 [ n(n+1)(2n+1)/6] = (2/3)n(n+1)(2n+1)

 Find th sum of series of odd squares in natural series

  S²_o = 1² + 3² + 5² + 7² ……… (2n – 1)²

The sum can be determined by finding out the mean common difference.

With two terms 1² + 3² = 10 =(2+d)   or  d = 8

With three terms  1² + 3² + 5² = 35 = 3 +3d or d = 10 + 2/3  = 8 + (2 +2/3)

With four terms 1² + 3² + 5² + 7² = 84 = 4 +6d or d = 13 +1/3 = 8 + 2 ( 2 + 2/3)

With n terms 1² + 3² + 5² + 7² ……… (2n – 1)²   , n + n(n-1)<d>/2  where the mean <d> = 8 +(n-2)[2 +2/3] = 8 + 8n/3 – 16/3 = (8/3) (n+1)

S²_o = n [ 1 + (n-1) <d>/2]

     = n [ 1 + (n-1)/2 [8/3 (n+1)]

     = n [ 1 +(4/3)(n² -1)] = (n/3) [4n² – 1]

 

 

 Find the sum of the given series  (1x2) + (2x3) + (3x4)……… [(n-1)n]  

       1²  =  1

      2²  =  2 + 1.2

    3²  =  3  + 2.3

    4²  = 4 + 3 x 4

    n² = n + (n-1)n

Summing up all the equations, ( 1² + 2² + 3² + 4² ……. n²) =( 1 + 2 + 3 + 4 ……. n)+ [(1x2) + (2x3) + (3x4)……… [(n-1)n]

We know , ( 1² + 2² + 3² + 4² ……. n²) = n(n+1)(2n+1)/6

( 1 + 2 + 3 + 4 ……. n) = n(n+1)/2

Hence , [(1x2) + (2x3) + (3x4)……… [(n-1)n] = n(n+1)(2n+1)/6  - n(n+1)/2

On simplification it reduces to n(n²-1) /3

 

 

The sum of series of successive cubes is equal to square of the sum of root numbers.

1³ + 2³ = 1+8 = 9 = 3² = (1+2)²

1³ + 2³  + 3³ = 1+8 +27 = 36 =6² = (1+2 + 3)²

1³ + 2³  + 3³ + 4³= 1+8 +27 + 64 = 100 =10² = (1+2 + 3 + 4)²

1³ + 2³  + 3³ + 4³ …… n³ =  (1+2 + 3 + 4…… n )² = [n(n+1)/2]²

This can be proved b considering this series as an arithmetic series with varying common difference.

If there are two terms 1³ + 2³  = 9 = 1 +(1+d₁ ) = 1 + (1+d) or d = d₁ = 7

With three terms 1³ + 2³  + 3³ = 36 = 3 + d₁ + d₂ = 3 + (1+d) +(1+2d) = 3 + 3d

                               d₁ + d₂ = 33 ; d₂ = 26 and d = 11 = 7 +4

With four terms  1³ + 2³  + 3³ + 4³ = 100 = 4+ d₁ + d₂ + d₃ = 4 + 6d

                                d₁ + d₂ + d₃  = 96 , d₃ = 63 and d = 16 = 7 + 4 +5

The mean d varies as d = 7+  [(n+1)(n+2)/2 – 3x4/2] = 7 + [ (n² +3n +2)/2 -6] = 1 +(n² +3n +2)/2 = (n² +3n +4)/2 where n ≥ 2

S³ = 1³ + 2³  + 3³ + 4³ …… (n-1)³  where  <d> =  (n² +3n +4)/2

 

   = n [ 1 + (n-1)/2 d] = n { 1 + [(n-1)/2][ (n² +3n +4)/2] = [n(n+1)/2]²

Sum of successive square numbers

S = 1² + 2² + 3² + 4² +……….   + n² , where n is the order of the number in the series which is equal to its value

These squares are grouped into series of odd squares 1² + 3² + 5² +……….   + (2n-1)²

And series of even squares 2² + 4² + 6² +……….   + (2n)²

Where n is the order of the odd/even number in the series with values  N_o – (2n -1) and N_e= 2n. It is noted that the sum of squares of all odd numbers from 1 to (2n-1) in natural series is 1/6 times the product of three successive numbers (2n-1)2n(2n+1)

 1² = (1x2x3)/6 = 1

1² +3² = (3x4x5)/6 = 10

1² +3² + 5² = (5x6x7)/6 =35 and so on.

1² + 3² + 5² +……….   + (2n-1)² = (2n-1)2n(2n+1)/6= (n/3)(4n² – 1)

In terms of the odd numbers  N_oH = 2n -1, 2n = (N_oH + 1) and 2n+1 = N_oH +2. In terms of N_oH ,the sum becomes So² = N_oH (N_oH +1) (N_oH +2)/ 6

 

Like wise the sum of all even numbers from2 to 2n in the natural series is 1/6 times the product of three successive numbers 2n (2n+1) (2n+2). For example,

 

2²  = (2x3x4)/6  = 4

2² +4² = (4x5x 6)/6 = 20

2² +4² + 6² = (6x7x8)/6 = 56

In general 2² + 4² + 6² +……….   + (2n)²  = 2n (2n+1)(2n+2)/6 = (2/3) n (n+1)(2n+1)

In terms of the even numbers  N_eH = 2n , n = (N_eH /2) , (n+1) = (N_eH +2)/2 and 2n+ 1 = N_eH +1. In terms of N_eH, the sum becomes S_e² =(1/6) N_{eH (}N_eH+1) ₍N_eH +2)

The sum  all squares in natural series is the sum of the sum of odd squares and sum of even squares S² = S_o² + S_e²

In terms of order the numbers \ (n/3) (4n² -1) + (2n/3)(n+1)(2n+1) = (n/3)(2n+1)(4n+1) . In terms of the actual value of the number N_H = 2n

S = N_H (N_H +1)( 2N_H +1)/6

 

There is yet another way to derive the general expression for the sum of squares in natural series S = 1² + 2² + 3² + 4² + ………… This can be considered as an arithmetic series with different common difference . By estimating the mean common difference , we can use the same formula used for arithmetic series with a common difference. The common difference may be different, but it varies gradually ,which can be estimated.

 

1² + 2² = 5 = 1 +(1+d₁) = 1 + (1+d) or d= d₁ = 3

1² + 2² + 3²  = 14 = 3+ d₁ + d₂ = 3 + 3d; d₂ + d₂ = 11 and d₂ = 8 and d = 11/3 = 3 +(2/3)

1² + 2² + 3²  + 4² = 30 = 4 + d₁ + d₂  + d₃ = 4 + 6d ; d₁ + d₂  + d₃ = 26  and d₃ = 15. d = 13/3 = 3+(2/3) +(2/3) = 4 + (1/3)

1² + 2² + 3²  + 4² + 5² =55 = 5 + d₁ + d₂  + d₃ + d₄ = 5 +10d ; d₁ + d₂  + d₃ + d₄ = 50 and d₄ =24 , d = 4 + (1/3) + (2/3) = 5.

It is noted that the mean d is gradually increasing  3 + (2/3) (n-2)

If there are n terms  1 +(1+d) + (1+2d) + (1+3d) …….. [1 +(n-1)d]

S = n + d[ 1+2+3……(n-1)] = n + n(n-1)d/2 = n[ 1 + (n-1)d/2]

Substituting the mean value of d, we get,

S = n { 1+ [(n-1)/2][ 3 + 2/3(n-2)]} = n { 1 + [(n-1)/2](9+2n-4)/3 } = n [1 +(n-1/2)(2n+5/3)] = n[(2n² +3n +1)/6 = n(n+1)(2n+1)/6   

  

 

1.solve a² + b²  = c² with the help of algebra where the factors of one of the square number is kown 

 With single variable a = α²

a²  = c^{2  -} b²  

α⁴ = (c-b)(c+b) .If c - b = α and c + b = α³ , which give c = α (α² +1)/2 and b = α (α^{2  -} 1)/2 . The solution then becomes (2α²)² +  [ α (α^{2  -} 1) ]² = [ α (α² +1) ]²

with two variables a = αβ      

α² β²  = (c-b) (c+b) . If c-b = α  and c+b = αβ² .By solving for b and c, we get a solution as

(2αβ)² + [α(β² – 1)]²  = [α(β² + 1)]²

The distribution of prime factors of a  to (c-b) and (c+b) may have different values ,however its product must be equal to the square of the prime factors of a. If If c-b = α²  and c+b = β² .By solving for b and c, we get a solution as

(αβ)² + (β² – α²⁾²  =  (β² + α²⁾²

(2) Solve a³ + b³ = c³ by using prime factor method

With single variable . let a = α^2. Then α^{6 =} (c – b) [(c-b)² + 3cb]. Under one possible distribution of prime factors (c-b) = α and [(c-b)² + 3cb].=  α² + 3cb = α⁵ or 3cb = α² (α³ -1) or b[= [α² (α³ -1)]/3c .Substituting this value in c-b ,  c -  [α² (α³ -1)]/3c = α  or  3c² – 3cα -[α² (α³ -1)] = 0. Solving this quadratic equation for c, we get

c =[3α ± √ (9α² + 12 α^5- 12 α² ) ]/6 = α/2 + (1/6) √ (12 α^3- 3 5² ) ] and b = c – α = - α/2 + (1/6) √ (12 α^{5 -} 3 α² ) ]. It gives its solution as (α ²)³ + [- α/2 + (1/6) √ (12 α^3- 3 α²)]³ = [ α/2 + (1/6) √ (12 α^5- 3 α²)]³ . Whatever may be the prime factors of a and its distribution among b and c, ,there will be integral solution. Ie., all a,b,c cannot have positive integers. This is called Fermat’s Last theorem.

With two varibales. Let a = αβ. Then  α³ β³ =   (c – b) [(c-b)² + 3cb].. If c-b = α² ,  then   [(c-b)² + 3cb]. = α⁴  + 3cb =  α β³ or b = [α (β³ – α³ )/3c. Substituting this value in c-b , c - [α (β³ – α³ )/3c.]^- = α²  or  3c² – 3c α² - α (β³ – α³ ) = 0. The solution of this quadratic equation becomes c = [3 α^{2 ±} √ 9 α⁴ + 12 α (β³ – α³ )]/6 . It is reduced to c = (α²/2) + (1/6) √  12 α β³ –3 α⁴ )  and b = c- α^{2 = -} (α²/2) + (1/6) √12 α β³ –3 α⁴ ) ^-.It gives the solution as (αβ)³ + [ (α²/2) + (1/6) √12 α β³ –3 α⁴)]³ = [(α²/2) + (1/6) √12 α β³ –3 α⁴)]³

(3) The prime factor method is application to solve a^x + b^y = c^z.,where the exponents are not same..S=olve a² + b³ = c^{3 .}for the integral values of a,b and c.

Let  a = =   α², then α⁴ = (c-b) [ (c-b)² + 3cb]. One possible distribution  is c-b = α and  (c-b)² + 3cb] = α² + 3cb = α³  or b = [α²  (α -1)]/3c  .Substituting this value of b in c-b  = α we get a quadratic equation 3c² – 3cα - α² (α -1) = 0, Solving for c, we get its two roots c =[(α/2) + (1/6)√ (12 α³ - 3 α² )]  and b = [ - α/2) + (1/6)√ (12 α³ - 3 α² )]]. The general form of the relation becomes ( α² )² + [(- α/2) + (1/6)√ (12 α³ - 3 α)]³ = [ (α/2) + (1/6)√ (12 α³ - 3 α)] . It is seen that when α = 1,7,19,37……. The root numbers become whole integers. When α = 7 ;  49² + 7³ = 14³

 (4) Solve a³  + b² = c² for integral values of a,b,and c.

 a³   = c²  - b²= (c-b) (c+b)

With single variable  a,  c-b =a and c+b = a² which give the value for b and c in terms of a.

c = a(a+1)/2 and b = a(a-1)/2. The relation in terms of a single variable a is a³ + [a(a-1)/2]² = [a(a+1)/2]²

 a=2 ; 2³ + 1²= 3²

a=3 ; 3³ + 3² = 6²

with two variables a = αβ; (αβ)³ = (c-b)(c+b) . One possible distribution of the prime factors is  c-b = α² β  and c+b = αβ^{2 .} It gives c = αβ ( β + α) / 2 and b = αβ (β - α) / 2. In terms of these variables, the relation becomes  (αβ)³ + [αβ (β - α) / 2]² = [αβ (β + α) / 2]²

α = 2,β = 3;  6³ + 3² = 15²

α = 2,β = 5; 10³ + 15² = 35²