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  .