Number theory

Number Theory of  a³ + b³  + c³  = d³

Abstract:

The number theory of cubical relations predicts that it is not possible to get  a cube equals to sum of two cubes, but it can be shown to  be equal to sum of three or more cubes. In this article few methods of generating such cubical relations and its general properties are studied. 

Key words:

Diophantine equation -Number theory- Fermat’s Last theorem, Beal conjecture

Introduction:

When a,b,c and  x are positive integers, then the like power relation  a^x + b^x  = c^x  is  not possible for  all values of x ≥3 (Fermat’s Last theorem), since in such relation any two members are proportional to the other member. If b and c are proportional to a, then  b^x = α a^x  and c^x   = β a^x  where the proportionality factors satisfy a condition 1 +α = β, which gives insignificant solutions due to the requirements α = p^x , β = q^x. However multi-power relation a^x + b^y = c^z  with at least one of the exponents is different from the other is possible, where a,b and c  have invariably a common prime factor (Beal conjecture).The missing common prime factor when the exponents become same can be taken as an indirect proof for FLT.

In the like power relations, it is found that  a number with an exponent 3 or more can be expressed as a sum of three or more numbers with the same exponent and such relations are generally called Diophantine equation. In this article the methods of generating a cube equals to sum of three cubes   and its associated mathematical properties are discussed. It gives yet another valid proof for FLT – in  a^x  + b^x  + c^x = d^x  if the number of members with like powers in the LHS reduces to 2, the integral  solutions disappear.

Diophantine  way
One can get a cube equals to sum of three cubes by many different ways using algebraic identities, which one can generate arbitrarily .

Let   (x+ ma)³  - (x- ma)³  = ` [(x-na)/b  + c]³  - [(x-na)/b  - c]³  be our algebraic identity of interest. When m ≠ n  and b >  1, the relation reduces to

m³a³  + 3max²  = c³  + (3c/b²) (x-na)² .

 If c = mab², then  x = (a/6n)[m²b⁶ + (3n² – m²)]. In  terms of m,n and b , the three independent variables, the most general relation becomes

{b[m² b⁶ +  (3n² – m²) + 6mn]}³  + [m² b⁶ +  (3n² – m²) – 6n² – 6mnb³]³  = {b[m² b⁶ +  (3n² – m²) - 6mn]}³  + [m² b⁶ +  (3n² – m²) – 6n² + 6mnb³]³.

The values of the independent variables can be so chosen so that one of the members becomes negative. Under the condition 3n²  + m²  + 6mnb³ > m² b⁶ , the given algebraic identity gives one cube equals to sum of three cubes.

m      n        b                 a³  + b³  + c³  = d³

 1      2        2                [45,126,147=174]³

1       3        2                [108,144,180=216]³

2       3        2                [63,486,513 = 630]

1       4        2                [174,177,207=270]³

2       4        2                [180,504,588 = 696]³

3       4        2                [ 57,1086,1095=1374]³

                    

There are quite a large number of algebraic expressions for a sum of two cubes equal with a sum of two other cubes. When one of the members gives negative value it is then transformed into a cube equal with a sum of three cubes. For example, the algebraic identity  (x-a)³  + (x+a)³  = (x-a +p n)³  + (x+a-q n)³  can be used to generate such relations by solving  for x and a,  and by giving convenient  values for p,q  and conditional value for n. When p = 9 and q = 1 and n = 3k where k = 1,2,3,4,5.......we have  x = [30a – 369)± 3 √ 36 a² – 540 a – 2343 ]/24 for  k = 1 and a =  [45 ± √ 4368 + s² ]/6  where s²  = 36 a² – 540 a – 2343. Few typical solutions are

n            s                a                   x            a³  + b³  + c³  = d³

3            11            56/3            28/3     [28,53,75=84]³

                                                   79/12   [145,179,267 = 303]³

              31            59/3            16/3     [38,43,66=75]³

                                             - 14/5          157/12   [79,245,357=393]³

              64           137/6         127/6     [5,76,123=132]³

                     71             71/3           16/3      [26,55,78=87]³

                                        -26/3         277/12   [7,317,525=561]³                                  

When the value of n is changed, it gives reducible relation with a common prime factor and consequently it reduces to the relation corresponding to n = 3.

Ramanujan’s way

In Ramanujan’s method, if p.q,r are so assumed as p + 3a²  = q + 3ab = r + 3b² = (a+b)²  and m, n are any two independent variables , then

n (mp + nq)³  + m (mq + nr)³  = m (np + mq)³  + n (nq + mr)³ . A particular case of the above theorem is

(3a² + 5ab – 5b²)³ + ( 4a² -4ab + 6b²)³  + (5a² – 5ab – 3b² )³  = (6a² – 4ab + 4b²)³ 

Some irreducible numerical relations are given  below

a          b                  a³  + b³  + c³  = d³

1         0                  [3,4,5=6]³

            2                 [7,14,17=20]³

                3                 [27,30,37=46]³

            4                 [54,57,63=84]³

2          3                 [ 3,36,37=46]³

3          1                    [27,30,37=46]³

           

When a= b, a sum of two cubes equals to sum of two other cubes can be obtained where the members in both sides will be identical. When b= 0, we get irreducible relation for a = 1 and reducible for its higher values, which is reduced to the irreducible one corresponding to a =1.It is interesting to note that the relation remains same when a and b are interchanged. When one of the members is made to be equal to zero, then the value of a for any integral value of b will become irrational and vice-versa, so that a cube cannot be shown to be equal to a sum of two cubes all with integers. Alternatively, it can be stated that it is not possible to get a number which when added with a cube cannot be expressed as another cube and as sum of two cubes simultaneously. 

Meyyappan’s way

In a³  + b³  + c³  = d³ , d is greater than a,b and c but less than a+b+c . Let d = c+n . then

a³  + b³  = 3cn(c+n) + n³ . In all the cubical relations, the difference between the sum of members in both sides of the relation will always be equal to some multiples of 6. i.e.,

a +b = 6m + n, which gives

a³   + b³ = (6m+n)³  - 3ab (6m+n)  =3c²n + 3cn²  + n³

Solving for c,

c = - (n/2)  + (1/6n) √ [ 288 m³n + n⁴  + 144 m² n²  + 24 m n² – 4abn(6m+n) }

      n           m          a+b           a³  + b³  + c³  = d³

     1            1            7                  [1,6,8=9]³

     1            2          13                  [3.10,18 = 19]³

      1             3          19                  [2,17,40 =41]³

      1             4          25                   not possible

     1             5          31                 [12,19,53=54]³

     1             6          37                 [14,23,70 = 71]³

      1             7          43                 [12,31,102=103]³

     1             8          49                   not possible

     1             9          55                   not possible

This method is well suitable to get a³  + b³  + c³  = (c+n)³  for a given value of ”a”  where n is an integer. It predicts that for every value of a (= m), there is a numeral relation  m³ + b³ + c³  = (c+1)³  and for any particular value of a, there would be one or more numeral relations a³  + b³  + c³  =(c+n)³ with different value of n.

When n = 1 and  a= m (m = 1,2,3,4,5,6....)

 b³  =3c²  + 3c + (1- m³ ) and   b = (6k + 1 – m)

 Solving for c ,   c =  ( -1/2 ) + (1/6) √  [12(b³ + m³) – 3]

 a (=m)     b          c            c+1(=d)

1               6           8              9

2             17         40            41

3             34        114          115

4             57        248          249

5             86        460          461

6            121       768          769

7            162      1190        1191

8            209      1744        1745

9            262      2448        2449

10          321      3320        332

It is interesting to note that as “a”  increases with a constant difference , all other members b,c and d  also increase proportionately. The general expression for this type of relation is

                 ^{n                                                                 n-1                                                                                n-1}

n³  + [ ∑ (6m)  - (n-1)]³  + [ 8n + ∑ 6 (3m-1)(n-m)]³   = [ 8n + 1 + ∑ 6 (3m-1)(n-m)]³ ]³

                 ^{m=1                                                          m=1                                                                               m=1}

For a given m (=a) , there  can be one or more relations with d-c = 1 , for example when m = 3.

m (=a)       b            c            c+1 (=d)

3              4             5               6

3             10          18             19

3             34         114          115

For same m (=a) , n ( =d – c) can have different selective values,

m= 3 ; n = 1;  [3,10,18=19]³ 

m= 3;  n = 9 ; [ 3,36,37=46]³

m = 7; n = 1 ; [7,162,1190=1191]³

m= 7 ; n = 3;  [7.14,17=20]³

m= 7 ; n=36 ; [ 7,317,528,561]³

Again it is found that the relation does not exist when either a or b becomes zero.When n= 0 or 1, the relation simply converts into 1³ = 1³ .  b=0  gives a condition   _mⁿ ∑ 6m =  (n-1) .It is possible only when n = 1 and m =0. For any other values of m and n   _mⁿ ∑ 6m ≠  (n-1). This is in support of FLT. Alternatively, it can be stated that it is not possible to get a number which when added with a cube cannot be expressed as another cube and as sum of two  different cubes simultaneously.

Properties of  a³  + b³  + c³  = d³                                 

1.Any four member sum of like power relations can be used to generate six member sum of like power relations. For example, [3,4,5 = 6]³   

[1,4/3,5/3 =2]³ and [ ¾,1,5/4 = 6/4]³  give  24³  + 9³  + 15³  = 18³ + 16³ + 20³

[1,4/3, 5/3 = 2]³  and [ 3/5,4/5,1 = 6/5)³ give 18³ + 20³ + 25³  = 30³ + 9³   + 12³

2.a^x + b^x  + c^x  = d^x has integral solutions for n = 2 and 3, provided a,b,c and d are not squares or higher powers simultaneously. . It predicts the non-existence of a⁶  + b⁶  + c⁶  = d⁶ .

The general form of a² + b²  + c²  = d²  is  a²  + a² [(k² – 1)/2k]² + a² [(k² –1)/2k]² [(k² +1)/2k]²  = a² [k²+ 1)/2k]⁴, where 1 +[(k² – 1)/2k]²  = [(k² + 1)/2k]²   If “a”  is  a square number , then (k² – 1)/2k = p², (k² –1)/2k] [(k² +1)/2k] =p² q²  which gives an unfathomable condition 1 + p⁴   = q⁴   , since the minimum difference between any two fourth power is 15. Similarly one can show that all the members in   a³ + b³  + c³  = d³  cannot be squares or cubes due to non-trivial condition 1 + p⁶  = q⁶

3. It can be shown that all the members cannot be primes simultaneously. a³  + a³ [(k² – 1)/2k]² + a³ [(k² –1)/2k]² [(k² +1)/2k]²  = a³ [k²+ 1)/2k]⁴ . If a = p_{1 ,} a³ (k² – 1)/2k]² = p₂³, a³ [(k² –1)/2k]² [(k² +1)/2k]² = p₃^{3 ,} then  the required condition p₁³ (p₃³ – p₂³ ) = p₂⁶ which predicts that p₂ cannot be a prime.