Pythagorean triples

 Pythagorean triples with same digital endings

 Sum of squares of two integers can be a square of an integer. Such a set of integers is called Pythagorean triple. The smallest known Pythagorean triple is (3,4,5). One can develop Pythagorean triples having same digital endings. For example in the case of (3,4,5) , we have first few Pythagorean triples with same digital endings

 3²  + 4²  = 5²

13² + 84² = 85²

23² + 264²= 265²

33²+ 544²= 545²  

43² + 924² = 925²

This set can be expressed with an algebraic relation

(10 n + 3)² + ( 50 n² + 30n + 4)² =  ( 50 n² + 30n + 5)²

mathematical puzzle

Fun with Mathematics

If p and q are positive and real numbers such that p²  +  q² = 1 . What are the maximum and minumum values of (p+q)  ?

Solution:

P²  = 1 – q²  = (1 + q) (1 –q). Let  1 + q = kp  and 1-q = p/k. By solving these two equations, we have

P = 2k/(k²+ 1)  and  q = (k² – 1)/(k² + 1). If p+q is maximum  p and q must be equal or k² -2k -1 = 0 . It gives k = (1 + √2) or p = q =(1 +√2)/(2+√2) It gives p+q = √2.

If p+q is minimum the difference between them must be maximum provided p or q cannot be greater than 1. 1+ q = p² and 1- q = 1. It is possible only when p = 1 and q = 0..So the minimum value of p+q = 1   

Discipline in Digital world

Discipline in digital world

Abstract: All numbers are having its own characteristics when mixed together in a relation, they show strict discipline which is curious to know. The disciplinary behavior of a group of numbers is quite amazing when exposed. Few inherent disciplines in the digital world are described in this article.

Key words: Number theory, Numeral relations- Triangular numbers

Introduction:

In numeral relations, two or more numbers are connected together with some mathematical operators When the numbers are varied under the same condition, the disciplinary behaviors of the numbers strike our mind. The inherent disciplines of a group of numbers in the digital world are exemplified with few simple numeral relations.

1.Sum of three squares of two successive numbers and its product

Sum of two squares of any two successive numbers is added with the square of its product always gives a square, the root number of which is one excess over the product number. This in turn is a factor for the sum of fourth powers of the same set of numbers.

1²  + 2² + 2²  = 3²  ;  1⁴  + 2⁴  + 2⁴  = 33 = 3 x 11

2² + 3²  + 6²  = 7² ; 2⁴ + 3⁴  + 6⁴  = 1393 = 7 x 199

3²  + 4²  + 12² = 13² ;  3⁴  + 4⁴  + 12⁴ = 21073 = 13 x 1621

4² + 5² + 20²   = 21² ; 4⁴ + 5⁴ + 20⁴   = 160881 = 21 x 7661

For any number n, it can be expressed as  n² + (n+1)²  +[n(n+1]²  =[ n(n+1) +1]²   and it is found that [n(n+1) + 1] is always a factor for  n⁴ + (n+1)⁴  +[n(n+1]⁴

2. Sum of three squares of any two numbers and its sum

Sum of squares of any two numbers when added with the square of its sum has a peculiar property . Half of the sum when added with the product of the two numbers taken gives the square of its sum 

   1²  +  12²  + 13²  =  2 x 157 ; 157 + 1 x 12 = 13²

    2²  + 11²  + 13² = 2 x 147 ; 147 + 2 x 11 = 13²

   3²  + 10²  + 13²  = 2 x 139 ; 139 + 3 x 10 = 13²

  4² +  9²  + 13²  = 2 x 133 ; 133 + 4 x 9 = 13²

For any two numbers m and n, we have

m²  + n² + (m+n)²  =  2( m² + n²  + mn)  ; m²  + n² + mn + (m x n) = (m+n)²

3. Sum of four different powers of any four successive numbers

  It is found that the sum of four different powers of any four successive numbers where both the numbers and the exponents are in ascending order has some curious properties

1 + 2² + 3³ + 4⁴  = 288 = 2 x 144 = 3 x 4² x 6

2 + 3² + 4³ + 5⁴ = 700 = 7 x 100 = 4 x 5²  x 7

3 + 4² + 5³ + 6⁴  = 1440 = 10 x 144 = 5 x 6² x 8

4  + 5²  + 6³ + 7⁴ = 2646 = 6 x 441 = 6 x 7² x 9

In general, n  + (n+1)² + (n+2)³ + (n+3)⁴  = (n+2) (n+3)² (n+5)

With any three successive numbers, we have

1 + 2² + 3³  = 32 = 2 x 4²

2 + 3² + 4³ = 75 = 3 x 5²

3 + 4² + 5³  = 144 = 4 x 6²

4  + 5²  + 6³ = 245 = 5 x 7²

In general, n + (n+1)²  + (n+2)³ = (n+1) (n+3)²

4 The sum of any three successive numbers when added with its mean gives the cube of that mean

 1 x 2 x 3 + 2 = 2³

2 x 3 x 4 + 3 =  3³

3 x 4 x 5 + 4 = 4³

In terms of n

n x (n+1) x (n+2) + (n+1) = (n+1)³

The square and cube of successive numbers are related with successive triangular numbers T_n

 which are nothing but the sum of natural numbers from 1 to n, 1,3,6,10,15,21,28,36,45,55,66,78………. [n(n+1)]/2

1²  = 1

2²  =  2 + 2(1)

3² = 3 + 2(3)

4² = 4  + 2(6)

n²  = n + 2(T_n-1)

The cubes have different correlation with the triangle numbers

1³ = 1

2³  = 2 + 6(1)

3³ =  3 + 6 (1+3)

4³  = 4 + 6(1+3+6) 

n³  = n + 6 x (∑ T_n-1)²

The fourth power of a number

1⁴  = 1

2⁴  =  2²  + 12 (1)                                                                        

3⁴ =  3²  + 12 (1+4)

4⁴   = 4²  + 12( 1+4+ 9)

In general, n⁴  = n² + 12  ^n-1∑ x²           

                                              ¹

More about Triangular numbers
Introduction:
The triangular number is a kind of figurate number that can be represented in the form of a triangular grid of dots or unit number 1, where the rows denotes its order n and the total counts up to that row gives the n th  triangular number. Simply said , the sum of the natural numbers up to n  is the n th triangular number and is denoted by Tn. which can be mathematically expressed as  Tn  =  n (n+1)/2. 
Properties of Triangular numbers;
The triangular numbers has many interesting properties. Some of them are 
1. All perfect numbers  are invariably triangular numbers. First few triangular numbers which are also perfect are 6 (T3) . 28 (T7), 496 (T31), 8128 (T127) 2. The natural series of triangular numbers are continuous alternate blocks of pair of odd and even numbers The sum of two numbers in any block gives even square while one from a block and the neighbouring  number in the adjacent block gives odd square. In all cases the sum of any two successive triangular numbers is a square of higher order among them Tn-1 + Tn  = n2  and Tn  + Tn+1  = (n+1)2. Subtracting one from the other we get an identity (n+1)2 – n2 =   Tn+1   -  Tn-1  = (2n +1) , which predicts that an odd number can be represented not only as a difference of  squares of two numbers with a difference of odd number but also as a difference of two triangular numbers whose orders are one above and below n. The difference between the fourth powers of any two successive numbers gives yet another identity  (n+1)4  - n4 = (2n+1) [ Tn+1 + Tn-1  + 2Tn] 3. A square can be expressed in terms of triangular numbers by three ways. n2  = Tn-1 + Tn = T(n2)  - T (n2 – 1)  = n + 2Tn-1’  It gives an identity T(n2) – Tn  = T(n2 -1) – Tn-1
4. With any successive three triangular numbers Tn-1, Tn ,Tn+1  , one can get an identity Tn-1 + T (n2) + Tn+1  = (n+1)2. A kind of discipline is found in such identity with 5,7,9------ successive triangular numbers. Tn-2 + Tn-1 + T (n2) + Tn+1 + Tn+2 = (n+2)2  + 1 Tn-3 +      Tn-2 + Tn-1 + T (n2) + Tn+1 + Tn+2 + Tn+3   = (n+3)2  + (1+4) Tn-4 + Tn-3 + Tn-2 + Tn-1 + T (n2) + Tn+1 + Tn+2 +Tn+3 + Tn+4 = (n+4)2 + (1+4+9) In general with m number of triangular numbers above and below the n th  triangular number, we get

 Tn-m  + Tn-m+1 + Tn-m+2 + ………………….  Tn2 + Tn+1 + Tn+2 + …………… Tn+m  = (n+m)2 +   (m-1) ∑ x2                                                                                                                                                     1                                       
          5 .  Any three triangular numbers with a common difference D in its order shows yet                                                                                                                                                  another numeral discipline
                   D = 1,  Tn-1 + (Tn)2  + Tn+1   = (Tn + 1)2       D = 2 , Tn-2  +  (Tn)2  + Tn+2  = (Tn  + 1)2 + (22  - 1)                                                                                                                                                                                                                                                                                                                                                 D = 3 , Tn-3  +  (Tn)2  + Tn+3  = (Tn + 1)2  + (32 – 1)                                                                                                                                                       In general, when any common difference D = m, Tn-m  +  (Tn)2  + Tn+m  = (Tn  + 1)2  + (m2 – 1) 6 Not only square but also any Power of any number cab be expressed in terms of triangular numbers. n3 = (Tn-1 + Tn) n  = n + 6 [ n-1 ∑ Tx ]                    1                         n4 = n2  + 4 Tn Tn-1 n5  = n3  + 6 n2   n-1 ∑ Tx    = n3  + 4n Tn-1  Tn
                                           
1 n6 = n2 + 12 [ n2 + 3 n-1 ∑ Tx ]  [n-1 ∑ Tx]                                                                                                                                                                                                                                                                                                                1                            1  7     The Pythagorean triples (a,b,c) satisfying the relation  a2  + b2   = c2  have some correlation with its corresponding triangular numbers. Substituting the values for a.b.c  in  terms of triangular numbers using the property (3) we have . T (c2) – T(b2) – T(a2)   =  T(c2 -1) – T(b2 – 1)  - T (a2 – 1)  = (ab)2                      Tc – Tb  - Ta   = Tc-1 – Tb-1 – Ta-1 a + b – c = 2 [ Tc-1 – Tb-1 – Ta-1]  It gives an additional condition required for Fermat’s assertion that  the relation an  + bn  = cn  where a,b,c are all positive integers  and the exponent n > 2 has no whole integral solutions.

T(cn)  -  T(bn)  - T(an)  = T(cn- 1)  - T(bn-1) – T(an- 1)

               8. Innumerable identities can be obtained for every Beal’s expression.  Considering the simplest                                                                                                                                                                                                                                                 irreducible form of a Beal’s relation  1+ 8 = 9 and using the property (3) we have                 T1 + T64 – 2T7 – T63  = T81 – 2T8  - T80             9. There are infinitely many square triangular numbers,  the first few positive numbers are               1(T1),36(T8) ,1225 (T49),41616 (T288) ,1413721(T1681) ,48024900 (T9800) ,1631432881( T57121),55420693056 (T332928)………… The order and the corresponding square triangular numbers can be determined very
simply from the knowledge of the two product factors of any known square triangular number.  Let a2 and b2  are the two product factors of a square triangular number Tx . If a and b are both odd and b>a, then the order of the square triangular number is b2, if a is even and b is odd, then its order will be          b2 – 1. One of the two product factors of the next higher order square triangular number would invariably be  (a+b)2 and the other product factor depends upon the odd-evenness of the given factors a,b. If both are odd the other product factors is (a+b+  smaller odd number in a,b)2, if one is odd and other is even, the other product factors is (a+b + even number in a,b)2.    Conclusion:  The identities with triangular numbers can be used as a tool  to study  relations in number theory 

 Logical proofs for Fermat’s Last Theorem

With the help of elliptic curve, Gerardin [4] derived a general expression for the equal sum of two, two fourth powers. It gives a way to prove FLT logically.

(a + 3a² – 2a³  + a⁵  + a⁷)⁴  + ( 1 + a² -2a⁴ -3a⁵  + a⁶ )⁴  =  (a - 3a² – 2a³  + a⁵  + a⁷)⁴  + ( 1 + a² -2a⁴ +3a⁵  + a⁶ )⁴

If one of the members is made to be equal to zero, the remaining relation will not be true for any integral values. The conditions  1 ± 3a + a⁴  + a⁶  = 2a²  due to disparity of odd-evenness and 1 = a² ( 2a² ± 3a³  - 1 – a⁴)  give only impractical solutions which can be considered as a confirmation of FLT.

 

FLT - Proof from Beal's relation

Proof of Fermat’s Last Theorem from Beal’s relation

Abstract:  Multi-power relations with reference to Beal are discussed with  numerical examples. Some additional properties of Beal’s conjecture are also identified to prove FLT

Keywords: Fermat Last Theorem, Beal Conjecture, Number theory

Introduction

Fermat’s last theorem says that integral solutions for a^x  +  b^y   = c^z  are not possible when all the exponents are same and greater than or equal to 3 . However it is possible if atleast one of the exponents is different. In such multi-power relations, according to Beal [1-3] the members when  all the  three members  a,b,c and the exponents x,y,z  are all positive integers such that (x,y,z) > 2, invariably have one or more common factors. Conversely the equation  a^x  +  b^y   = c^z has no solutions for any positive integers a,b,c,x,y and z with (x,y,z)  > 2  for any set of co-prime (a,b,c).The three members like or unlike power relations described by  the Fermat’s assertion and Beal conjecture respectively  have some similarities and differences.

The observed similarities are (i)  All the members can be even, then the relation will be reducible. All the members cannot be odd due to non-conservation of oddness (ii) One or two members but not all the three may be square or higher power but all the members cannot be square or with the same higher power, (iii) all the three members cannot be simultaneously primes (iv) FLT and Beal conjecture are mathematically true with fractional and rational or complex numbers, and (v) In irreducible form, both the FLT and Beal conjecture do not have a common prime factor among the members

Among the differences the most significant are (i) In a^x  + b^y  = c^z , FLT requires that a,b,c cannot be positive integers when the exponents x,y,z are same and greater than 2, but they can be in Beal’s expression  when the exponents are different (ii) In FLt, a + b will always be greater than c. but it can be greater, lesser or even be equal depending upon the value of the exponents

In this paper Beal’s expression in the form of aⁿ + aⁿ  = 2^m is studied to investigate more properties of Beal’s conjecture and by using such properties FLT is proved.  

 Multi-Power Relation of type aⁿ + aⁿ  = 2^m

We know that aⁿ + aⁿ  =  2 aⁿ, To make 2aⁿ  to be a power of a number the only possibility is “a” itself must be  equal to 2^x . Under this condition it gives  2^nx + 2^nx = 2^nx+1 , where n and x have all positive whole integral values. For different values of x we have

2ⁿ  +   2ⁿ    =  2^{n + 1}

4ⁿ    +  4ⁿ    =  2^{2n + 1}

8ⁿ    +  8ⁿ   =  2^{3n + 1}

16ⁿ  +  16ⁿ  = 2^{4n + 1}

32ⁿ   + 32ⁿ   =  2^{5n + 1}

All of them have common prime factor 2 with which they can be reduced to an irreducible form 1 + 1 = 2, which is a fundamental requirement for this type of relations. Any power of 2 can be a common multiplier to generate multi-power relations with bigger numbers. For example  2ⁿ  +   2ⁿ    =  2^{n + 1} gives  2²ⁿ⁺¹  + 2²ⁿ⁺¹  = 4ⁿ⁺¹ when all the members are multiplied by 2ⁿ⁺¹ .

This method of getting multi-power relations, gives an ingenuous explanation for Fermat’s Last Theorem. Sum of two identical cubes can be expressed as a square or fourth or fifth power of a number, but never as a single cube or with any exponents of multiples of three.

(2x²)³ + (2x²)³ =(2²x³)²      (2x⁴)³  + (2x⁴)³  = (2x³)⁴                (2³x⁵)³ + (2³ x⁵)³  = (2² x³)⁵

2³ + 2³ =  4²                                       32³ + 32³   = 16⁴                                  8³  +  8³  =  4⁵

8³ + 8³  =  32²                                  162³ + 162³ = 54⁴                             256³  + 256³  = 32⁵

18³  + 18³  = 108²                        512³  + 512³ = 128⁴                         1944³ + 1944³  = 108⁵

When the exponent of the sum is in the form of  3n+2, the Beal, expression of this type is

[2²ⁿ⁺¹ xⁿ ]³  + [2²ⁿ⁺¹ xⁿ ]³ = (2² x³)^{3n +2}, and it is (2ⁿ xⁿ)³ + (2ⁿ xⁿ)³  = (2x³) ³ⁿ⁺¹  when the exponent is in the form of 3n + 1. It is noted that the exponent of the sum will never be in the form of 3n.. To be true, it should satisfy the relation  (2ⁿ xⁿ)³   + (2ⁿ xⁿ)³   = (2^m  x)³ⁿ which necessitates  the condition 2^{3n + 1}  = 2^3mn

No integer values for n and m satisfy this condition, which substantiates the Fermat’s last Theorem without any further argument.

FLT and Beal's Conjecture

Algebraic Proof for Fermat’s Last Theorem and Beal’s Conjecture

Abstract:

There are many proofs to prove Fermat’s Last Theorem as well as Beal’s Conjecture. Here is yet another proof, much simpler than any other proofs.

Keywords:

Fermat’s Last Theorem- Beal’s Conjecture- algebraic proof – numerical examples

Algebraic Proof for Fermat’s Last Theorem and Beal’s Conjecture

Introduction:

According to Beal all three member multi-power relations  a^x  +  b^y   = c^z  , where a,b,c and the exponents x,y,z  are all positive integers such that (x,y,z) > 2, the base numbers a,b,c  will invariably have one or more common factors. Conversely the equation  a^x  +  b^y   = c^z has no solutions for any positive integers a,b,c,x,y and z with (x,y,z)  > 2  for any set of co-prime (a,b,c). In such three members like or unlike power relations both the Fermat’s assertion and Beal’s conjecture have some similarities and differences.

The observed similarities are (i)  All the members can be even, then the relation will be reducible (ii) All the members cannot be odd due to non-conservation of oddness (iii) One or two members  may be square or higher power but all the members cannot be square or with the same higher power, (iv) All the three members cannot be simultaneously primes (v) FLT and Beal’s conjecture are mathematically true with fractional and rational or complex numbers, and (vi) In irreducible form, both the FLT and Beal’s conjecture do not have a common prime factor among the members.

Among the differences the most significant are (i) In a^x  + b^y  = c^z , FLT requires that a,b,c cannot be positive integers when the exponents x,y,z are same and greater than 2, but they can be in Beal’s expression, when the exponents are different (ii) In FLT,( a + b) will always be greater than c. but it can be greater, lesser or even be equal depending upon the value and position of the exponents in the multi-power relation. These differences give an explanation why Beal’s conjecture is allowed and Fermat’s last theorem is not for exponents > 2 .

In this paper both the Fermat’s last theorem and Beal’s conjecture  are proved with an algebraic method which is much simpler than any other methods. One and the same expression admits multi-power relations in accordance with Beal’s conjecture and forbids like-power relations in agreement with Fermat’s last theorem.

Algebra of three member power relation

In all the three member like  aⁿ + bⁿ  = cⁿ , and  unlike  (a^x + b^y  = c^z)   power relations, one can establish that  any two members are proportional to the remaining member.

a^x  = c^z – b^y  = (c^z/2 – b^yl2 )(c^z/2 + b^y/2 )

 If c^z/2 – b^yl2  = a^x/2/k  then  c^z/2 + b^y/2 = k a^x/2 , where k is a proportional factor

It gives c^z  = a^x [ (k² + 1)/2k]²= a^x β²  and b^y  = a^x [(k² – 1)/2k]² = a^x α². It clearly indicates that all the three member power relations have the condition  1 +    α²  =  β² as a common feature.  This unique condition clearly predicts the mathematical reason why whole integral solutions are not permitted when all the exponents are same and greater than 2, and permitted when the exponents are different. The difference between the  FLT and Beal’s equation lies in the successful formation of a^x β²  and a^x α² into  single power.

In fact this mutual dependence among the members provides them a common factor ,which is usually one of the members of the Beal’s equation. If we are interested whole integral solutions . the proportional factors α², β² must not only be positive integer, but also its product with the common power  a^x must be expressible as another power. When all are supposed to be expressed  in the same power, the proportional factor must also be  expressible in the same power.   since 1 + αⁿ   = βⁿ is not possible for any integral  values of α and β  when n >2, aⁿ + bⁿ = cⁿ    is possible only for n = 2 and becomes impossible when n > 2,which  is in accordance with FLT. When  the proportional factors obeying the unique condition are expressible as different power and  give different power when multiplied with the common power it becomes a multi-power relation  so-called Beal’s equation.

Forbidden Fermat and allowed Beal

If a,b,c,m and n  are all positive integers, aⁿ + bⁿ  = c^m  is allowed as Beal’s  relation when m ≠ n, and forbidden due to FLT when m = n.

(a^n/3)³ + (b^n/3)³ = c^m

P³ + q³ =  (p+q) (p² –pq +q²) =  c^m , where  p = a^n/3 and q = b^n/3. P and q to be whole integers, a and b must be cubes or n must be some multiples of 3. If p + q = α c (for m >n ) and p+q = c/α (for m <n) where α > 1  may be a fraction or whole integer but not complex .In the case of m >n,  p² –pq +q²  =   c^m-1/α  and  3q² - 3αcq + α²c² – c^m-1/α = 0 by eliminating p. Its solution is q = (αc/2) ± (αc/6)                    √[( 12 c^m-3/α³) – 3]. When m <n ,α <1 and hence  p² –pq +q²  =  α c^m-1  and  3α²q² - 3αcq +c² –α³ c^m-1 = 0 by eliminating p. Its solution becomes q = (c/2α) ± (c/6α) √[( 12α³  c^m-3) – 3].

This solution gives adequate explanation for the FLT and Beal’s conjecture. “q” to be whole integer, the terms inside the square root must be either z.ero or a square number s². For positive roots                4 c^m-3/α³  > 1 and for integral roots c^m-3/α³  = 3, 7,12,19,28,39,52,67,........ so that c is related to α.

When m = 3,  1< α < 1.6, the solutions for p,q and hence a,b are irrational for all fractional values of α  The real solutions are obtainable only  when α is whole integer. When  α ≥ 1.6 , the roots become imaginary . The non-existence of real roots is the clear indication of FLT. However if m is different form n, the solutions are obtainable. For example, when n =3, m = 4  , 12c/α³ = 7, then α = 6, c = 126, which gives 126³ + 630³   = 126⁴, when n = 3 , m = 5, 12c/α³ =28, then α = 21 , c = 147 which gives  4116³  = 147⁵  + 1029³ . Solutions are found when n=3,m = 2, 12/cα³ = 4, α = 1,c=3  which  gives 1³ + 2³ = 3^2.,it is 2³  + 2³  = 4²  for α = 1 , c = 4. For various values of α and c, one can get innumerable relations.

                m = 2                                                        m=4                                                       m=5

α              c            [a,b]³ = c²                                        α           c        [a,b]³ = c⁴                     α       c        [a,b]³ = c⁵

2             32          {8,8]³ = 32²                        2            2       [2,2]³=2⁴                      3        3       [3,6]³  = 3⁵

3            108         [18,18]³ =108²                  3            9      [9,18]³ = 9⁴                             4         4       [8,8]³ = 4⁵

 2                     24         [4,6]³  =24²                                     4           16     [32.32]³ = 16⁴                      12      24      [96.192]³ =24⁵

 7           588        [14,70]³ = 588²                  5           35    [70,105]³ =35⁴               16      32    [256.256]³ =32⁵   

m=6,,9,12......3n is not possible due to FLT.When n = 3,m = 7 ,12c⁴ /α³ =4, α = 27, c = 9 which gives  81³  +  162³  = 9⁷

In terms of α and c, the multi-power relation can be expressed as

[(αc/2) – (αc/6)√ (12c^m-3 /α³  - 3)]³     [(αc/2) + (αc/6)√ (12c^m-3 /α³  - 3)]³  = c^m                         

It shows that (p,q,c) and hence (a,b,c) has a common prime factor and this is in concurrence with Beal’s conjecture. Reducing the expression by dividing with the common factor and keeping α = 1, one can get the irreducible form of Beal’s relation. It is exemplified for a typical set of Beal’s relation  a³   + c⁴  = b³

[(1/2) + (1/6)√(12 c – 3)]³  -   [(1/2) + (1/6)√(12 c – 3)]³  = c

 c                    irreducible à reducible Beal’s relation

7                      2³  - 1³   = 7   à   x 7³  = 7³  + 7 ⁴  = 14³

19                   3³  -  2³  = 19 à   x 19³  = 38³ + 19⁴ = 57³

37                   4³ –  3³ =  37 à  x 37³   = 111³  + 37⁴  = 148³

61                  5³   -  4³    = 61 à x 61³  = 244³ + 61⁴  = 305³

This set of relations can generally be expressed as

(3n²   + 3n + 1) ;      (n+1)³  - n³= 3n²   + 3n + 1  à[(n)( 3n²   + 3n + 1)]³ + [( 3n²   + 3n + 1)]⁴ =  [(n+1)( 3n²   + 3n + 1)]³  .  

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.