A Proof of Fermat's Last Theorem using an Euler's

Asian Research Journal of Mathematics 6(3): 1-24, 2017; Article no.ARJOM.36405 ISSN: 2456-477X

A Proof of Fermat’s Last Theorem using an Euler’s Equation P. N. Seetharaman1 1 Energy Conservation Cell, Research

∗

& Development, Engineer (Retired), Tamil Nadu State Electricity Board, Tamil Nadu, India. Author’s contribution

The sole author designed, analyzed and interpreted and prepared the manuscript. Article Information DOI: 10.9734/ARJOM/2017/36405 Editor(s): (1) Junjie Chen, Department of Electrical Engineering, University of Texas at Arlington, USA. Reviewers: (1) Francisco Bulnes, Technological Institute of High Studies of Chalco, Mexico. (2) Taxiarchis Papakostas, Technological Educational Institute of Crete, Greece. Complete Peer review History: http://www.sciencedomain.org/review-history/21048

Received: 27th August 2017 Accepted: 11th September 2017

Original Research Article

Published: 18th September 2017

Abstract Fermat’s Last Theorem states that there are no solutions to xn + y n = z n for n ≥ 3 and x, y, z non-zero integers. Fermat wrote down a proof for n = 4 [1]. In 1753, Lenohard Euler (1707–1783) wrote down a proof of FLT for the exponent n = 3 [1]. Since any integer n ≥ 3 is divisible by an odd prime number or by 4, it is sufficient to prove FLT for n = p, an odd prime > 3. We prove the theorem for p ≥ 5. We consider x3 + y 3 = z 3 and sp + tp = up , where p is any prime > 3. Without loss of generality it is enough to assume that both x and y as non-zero positive integers; therefore z 3 will be a non-zero positive integer, but both z and z 2 will be irrational in the first equation. We hypothesize that there exist positive integers, s, t and u in the Fermat’s equation sp + tp = up and bring a contradiction. We have created by trial and error method two equivalent equations to Fermat’s equations. x3 + y 3 = z 3

and

sp + tp = up

through parameters a, b; c, d; e and f given by

*Corresponding author: E-mail: [email protected];

Seetharaman; ARJOM, 6(3): 1-24, 2017; Article no.ARJOM.36405

)2 √ ( √ )2 ( ( √ )2 a u+b a − b l5/3 √ √ + = e 71/3 k5/3 + f 2n/2 23n/2 ( √ )2 ( )2 ( √ √ )2 √ c xt + d √ + c y − d l7/3 = e − f z 75/3 k7/3 respectively. The values of a, b; c, d; e and f in terms of known values are worked out in the detailed proof. Also we can obtain the equivalent values of 2n , 7k2 and l2 from these equations where 2n = 7k2 + l2 . (The validity of the equivalent equations has been discussed in detail in Annexure–A). Solving the above two equivalent equations using the above Euler’s equation, we arrive at the result stu = 0, thus proving the theorem. Use of the Euler’s equation leads us to this elementary proof. Keywords: a, b; c, d; e, f parameters in the equivalent equations to Fermat’s equations x3 + y 3 = z 3 and sp + tp = up ; 2n = 7k2 + l2 . 2010 Mathematics Subject Classification: 11A–xx.

1

A Few Previous Works on Fermat’s Last Theorem 1) The French mathematician Pierre De Fermat (1601–1665), who conjectured that the equation xn + y n = z n has no solutions in positive integers x, y, z if n is a positive integer ≥ 2, had seemed to have proved the theorem only for n = 4 by ‘Infinite Descend Method’ [1]. 2) In 1753, Leonhard Euler wrote down a proof of FLT for n = 3 [1]. Gustav Peter and Adrien Marie Legendre proved the case for n = 5 in 1825. In 1839, Fermat’s equation for exponent 7 also yielded to elementary methods through the heroic efforts of Gabriel Lame. 3) Around 1820, Sophie Germain (1776–1831) proved that if l is a prime, and q = 2l + 1 is also prime, then Fermat’s equation xl + y l = z l has no solutions (x, y, z), with xyz ̸= 0(mod l). Germain’s theorem was the first really general proposition of FLT, unlike the previous results which considered the Fermat’s equation taking the values of n, one at a time. 4) The work of Ernst Edward Kummer (1810–1893) marked the beginning of a new era in the study of FLT. For the first time, sophisticated concepts of algebraic number theory and theory of L-functions were brought to bear on a question that had until then been addressed only with elementary methods and showed FLT is intimately tied to deep question on class numbers of cyclotomic fields which are still an active subject of research [1], [2], [3], [4]. 5) In 1955, Yutaka Taniyama (1927–1958) posed some questions regarding elliptic curves additional investigations by Andre Weil (1906–1998) and G. Shimura resulted in the Shimura – Taniyama Weil conjecture. In 1986, a connection between this conjecture and FLT was discovered [5]. 6) Starting around 1970, a series of mathematicians started to get glimpses of a strange connection between elliptic curves and Fermat’s last theorem. In 1986, Gerhard Frey suggested that there was a correspondence between FLT and elliptic curves [4], [5], [6]. 7) Many eminent mathematicians around the world have contributed to FLT and we have furnished only a few milestones [7], [8]: Finally Andrew Wiles (1953– ) and Taylor – Wiles solved FLT completely in 1995. (Andrew Wiles : Modular elliptic curves and Fermat’s Last Theorem. Annals of mathematics 141 (1995), No. 3, 443–551). But he has used highly complicated mathematical developments [9], [10]. However we have given an elementary proof.

2


2

Introduction

If n > 3, then 2n can be represented in the form 2n = 7k2 + l2 , where k and l are odd positive integers [11]. Figure below shows the first nine solutions, which can easily be found by guessing. n k l

3 1 1

4 1 3

5 1 5

6 3 1

7 1 11

8 5 9

9 7 13

10 3 31

11 17 5

(Courtesy: Problem solving strategies Arthur Engel (Springer) pages 126–127).

3

Assumptions 1) Let us hypothesize that there exist non-zero positive integers s, t and u in Fermat’s equation sp + tp = up

(1)

where p is any prime > 3. We prove Fermat’s last theorem in the above equation by the method of contradiction. Clearly gcd(s, t, u) = 1. 2) Let us choose x, y and z 3 as square-free, non-zero positive integers in the Fermat’s equation x3 + y 3 = z 3 3

(2) 3

where gcd(x, y, z ) = 1. We also choose all x, y, z such that each is pairwise relatively prime to s, t, u and l where l is connected by the Euler’s equation 2n = 7k2 + l2 . (Discussed later on). 3) We have created equivalent equations by trial and error method to the above two Fermat’s equations through parameters a, b; c, d; e and f . Also we can find the equivalent values of 2n , 7k 2 and l2 connected by the Euler’s equation 2n = 7k2 + l2

(3)

from the equivalent equations to Fermat’s equations, where n ≥ 3, k and l are odd positive integers. Let us choose k and l such that they are > 1, both of them are not perfect squares and also they are not perfect cubes. What are the parameters and how the equivalent values are established through theses parameters have been explained fully in the detailed proof and in Annexure – (A), and the Euler’s equation 2n = 7k2 + l2 has been discussed fully in Annexure – (B). 4) Choose l as a square-free integer, co-prime to u. This could be possible for infinitely many values of l. (since l can be chosen infinitely many ways, for infinitely many values of n ≥ 3 in the equation 2n = 7k2 + l2 ). (Discussed in detail in Annexure – (B)) (pages 22 to 24). 5) Let us choose n as any odd integer ≥ 3 in the Euler’s equation 2n = 7k2 + l2 for the sake of convenience in this proof. By substituting the equivalent values of 2n , 7k2 and l2 that we get from the equivalent equations to Fermat’s equations, in the Euler’s equation 2n = 7k2 + l2 , we obtain another equation, called equation (B) in this proof. Our aim is to compute the rational terms on both sides in equation (B), in which we get some rational terms, either free from any factor involving square-root (radicals) or having some squareroot term as a factor. In equation (B), we are facing with rational terms multiplied by

3


√ √ √ √ 1/2 n/2 3 3n/2 3 3 p p 3 p any y u) xyz 3 sp up ; xytu; √ radicals √ z s t ; xyz s l; t (2 x + 2 √ one of the lz 3 sp tp up ; yz 3 sp up ; lu. √ We consider all these terms as irrational, as proved below: For example, suppose xyz 3 sp l is an integer equal to A (say), then A2 = xyz 3 sp l, which is impossible according to unique factorization theorem since x, y, z 3 are square-free integers, co-prime to each other and relatively prime to s and l. [see Annexure – (B)]. 6) If we choose n as an even integer, then by multiplying the following terms in the II term in L.H.S. of equation (B), {

−

√

( )√ √ l5/3 sp u xtl7/3 2n/2 · xsp tp l5/3 u(p+1) 23n/2 y 3

( ) √ √ } × 2n/2 x3 + 23n/2 y 3 u (−1) xytp+1 (− up )

(4)

we get { ( )( )( )√ } − 2n 2n/2 x3 + 23n/2 y 3 u x2 y 2 l4 sp t(p+1) u(p+1) t

(5)

and we have to include this term as a rational term, if t is a perfect square-integer and our proof will be restricted to only for t as a square integer. In order to avoid this situation and prove the theorem for all positive values of t, we consider both 2n/2 and 23n/2 as irrational, by choosing n as an odd integer in the Euler’s equation 2n = 7k2 + l2

(6)

Also since gcd(s, t) = 1, both s and t cannot be even integers. Hence we may consider, without loss of generality, t as an odd integer. Hence t1/2 (2n/2 x3 + 23n/2 y 3 u) cannot be a rational number, while n is an odd integer. √ ( ) ( ) Instead of 2n/2 x3 + 23n/2 y 3 u , if it is replaced by − 2n+1 x3 y 3 u (in the above said computation of rational term) then we get a rational term multiplied by the radical term √ xytu, which we define as irrational by choosing both x and y as square-free integers, gcd(x, y) = 1, and both x and y are pair-wise relatively prime to t and u. Since we are proving FLT only in the Fermat’s equation sp + tp = up

(7)

where p is any prime > 3, for all possible values of s, t and u as non-zero positive integers, we have got the liberty of choosing x, y, z 3 as square-free integers, each relatively prime to s, t, u and l, in order to avoid any radical factor in the evaluation of rational terms in equation (B) in this proof, and we also have the liberty of choosing n as an odd integer, k and l as square-free and cube-free integers in the Euler’s equation 2n = 7k2 + l2

(8)

√ 7) Consider the first term in R.H.S. of equation (B) which is {(bcd)(x2 y 3 z 3 t)}. While it is multiplied by the concerned terms as stated under the computation of rational terms on equation (B) as given in the detailed proof and on substituting for (bcd) we get (see the

4


detailed proof) ( √ )( )2 ( )√ x2 y 3 z 3 t ul5/3 − 1 71/3 51/3 z − 1 2n/2 xsp tp l5/3 u(p+1) ×

(√

2n/2 x3 ul5/3 −

√

) √ √ 2n/2 x3 − u 23n/2 y 3 l5/3 + 23n/2 y 3 u

( √ √ √ √ ) × 75/3 k7/3 sp l7/3 y − 75/3 k7/3 l7/3 xsp tp+1 + 75/3 k 7/3 sp tp y − tp xt ( ) √ × xtl7/3 + y − 2 xytl7/3

(9)

[( √ )( )( )√ ( √ ) i) If we multiply the terms x2 y 3 z 3 t 1 −1 2n/2 xsp tp l5/3 up+1 −u 23n/2 y 3 l5/3 ( )] { ( )√ } √ √ )( − tp xt xtl7/3 , the we get − 2n x4 y 3 tp+2 l4 up+1 z 3 sp tp ii) Alternatively, if we multiply the following terms ( √ )( )( )√ x2 y 3 z 3 t − 2ul5/3 or 1 − 1 2n/2 xsp tp l5/3 u(p+1) )( (√ √ )( 23n/2 y 3 u − tp xt y

or

−2

) √ xytl7/3 ,

(10)

then we get [ ( ) ( )] √ √ √ − 2n+1 x3 y 4 t(p+1) u2 l2 lz 3 (stu)p (or) − 2n+1 l2 x3 y 3 t(p+1) u t(p+1) xyz 3 sp up (11) √ [by multiplying (−2ul5/3 ) by y (or) 1 by (−2 xytl7/3 )]. √ √ √ In the above terms under (i) and (ii) we consider all z 3 sp tp , lz 3 (stu)p and xyz 3 sp up as irrational, since we have chosen x, y and z 3 as a square-free integers with gcd(x, y, z 3 ) = 1 and each is pairwise coprime to s, t, u and l [Discussed in detail in Annexure–(B)]. This justifies why we have chosen x, y, z 3 as square-free integers, gcd(x, y, z 3 ) = 1 and all x, y, z 3 are each relatively prime to s, t, u and l. One more example is given under item (8) below: 8) If we multiply the following terms in the I terms on L.H.S. of equation (B) [see the detailed proof] {√ ( )√ ( )√ ( ) √ sp u − y 2n/2 xsp tp l5/3 u(p+1) 2n/2 l5/3 x3 2n/2 x3 ul5/3 3 ×

(

)(√ )( √ )} − xt tp l7/3 − up

(12)

then we get ( ) ( ) ( ) √ √ −3 x6 y 2n l2 sp tp up+1 l5/3 l5/3 u = − 3x6 y 2n l4 sp tp u(p+1) lu,

(13)

√ in which we consider lu as irrational, if we choose l as a square-free integer and relatively prime to u where 2n = 7k2 + l2 . (Discussed in detail in Annexure–(B)). Examples:

211 = 7 · 172 + 52 = 2048 13

2

= 7 · 23 + 67 = 8192 2

2

(14) (15)

5


217 = 7 · 892 + 2752 = 131072

(16)

19

= 7 · 271 + 101 = 524288

(17)

21

= 7 · 457 + 797 = 2097152

(18)

2 2

2 2

2 2

223 = 7 · 2872 + 27952 ; 225 = 7 · 9672 + 51972 ; 227 = 7 · 40492 + 44112 and so on. [See Annexure–(B), to see how they are derived]. Note: If k = 1, then Euler’s equation gets reduced to Ramanujan – Nagell equation 2n = 7 + l2 which has exactly five solutions given by { } (n, l) = (3, 1); (4, 3); (5, 5); (7, 11); (15, 181) (19) The advantage of using Euler’s equation with k ̸= 1, over Ramanujan – Nagell equation is that the former equation has got √ infinitely many solutions and we may choose l as an odd integer infinitely many ways such that lu is irrational, whatever be the value of u as a positive integer. [Discussed in detail in Annexure–(B) (pp. 22–24)]. Proof. Let the equations )2 √ ( √ )2 ( )2 ( √ a u+b a − b l5/3 √ √ + = e 71/3 k5/3 + f 2n/2 23n/2 and

( √ )2 ( )2 ( √ √ )2 √ c xt + d √ + c y − d l7/3 = e − f z 75/3 k7/3

(20)

(21)

be equivalent to the Fermat’s equations x3 + y 3 = z 3

(22)

sp + tp = up

(23)

and

respectively through parameters a, b; c, d; e and f . Therefore √ √ a u + b = 2n/2 x3 √ √ a − b l5/3 = 23n/2 y 3 √ √ c xt + d = 75/3 k7/3 sp √ √ √ c y − d l7/3 = tp √ √ e 71/3 k5/3 + f = z 3 √ √ e − f z = up

(i) (ii) (iii) (iv) (v) (vi)

Solving the identities (i) and (ii); (iii) and (iv); (v) and (vi) simultaneously, we get the values of the parameters as worked out below: √ Multiplying both sides of (i) by l5/3 we get √ √ √ a l5/3 u + b l5/3 = 2n/2 l5/3 x3 (24)

6


√ √ (ii) gives a − b l5/3 = 23n/2 y 3 Adding, we get (√ ) (√ ) √ a l5/3 u + 1 = 2n/2 l5/3 x3 + 23n/2 y 3 ∴

a=

(√

2n/2 l5/3 x3 +

√

23n/2 y 3

)(√

ul5/3 − 1

(25) )/( ) ul5/3 − 1

(26)

√ √ (i) gives a u + b = 2n/2 x3 . √ Multiplying both sides of (ii) by u, we get √ √ √ a u − b ul5/3 = 23n/2 y 3 u

(27)

Subtracting, we get (√ ) (√ ) √ b ul5/3 + 1 = 2n/2 x3 − 23n/2 y 3 u

(28)

∴

b=

(√

2n/2 x3 −

)(√ )/( ) √ 23n/2 y 3 u ul5/3 − 1 ul5/3 − 1

√ Multiplying both sides of (iii) by l7/3 , we get √ √ √ c xtl7/3 + d l7/3 = 75/3 k 7/3 l7/3 sp √ √ √ (iv) gives c y − d l7/3 = tp . Adding, we get (√ √ ) (√ 5/3 7/3 7/3 p √ p ) c xtl7/3 + y = 7 k l s + t ∴

c=

(√

75/3 k7/3 l7/3 sp +

) √ )(√ √ )/( 7/3 tp xtl7/3 − y xtl −y

√ √ Multiplying both sides of (iii) by y and (iv) by xt we get √ √ √ c xyt + d y = 75/3 · k5/3 sp y √ √ √ c xyt − d xtl7/3 = xt(p+1) Subtracting we get (√ ) √ √ ) (√ 5/3 7/3 p d xtl7/3 + y = 7 k s y − xt(p+1) ∴

d=

(√

75/3 k7/3 sp y −

Multiplying both sides of (v) by

√

)√ ) √ )/( 7/3 xt(p+1) ( xtl7/3 − y xtl −y

(30)

(31) (32)

(33) (34)

(35) (36)

√

z, we get √ √ e 71/3 k 5/3 z + f z = z 2

√ √ (vi) gives e − f z = up . Adding, we get (√ ) ( √ ) e 71/3 k5/3 z + 1 = z 2 + up ∴

(29)

( )/( ) √ )(√ e = z 2 + up 71/3 k 5/3 z − 1 71/3 k5/3 z − 1

(37)

(38) (39)

7


(v) gives √ √ e 71/3 k5/3 + f = z 3

(40)

√ Multiplying both sides of (vi) by 71/3 k5/3 , we get √ √ √ e 71/3 k5/3 − f 71/3 k5/3 z = 71/3 k5/3 up

(41)

Subtracting we get (√ ) (√ ) √ f 71/3 k5/3 z + 1 = z 3 − 71/3 k5/3 up

(42)

∴

f=

(√

z3 −

√

71/3 k5/3 up

)(√

71/3 k5/3 z − 1

)/( ) 71/3 k 5/3 z − 1

(43)

From (i) and (ii) we get ( √ ) ) ( √ a u+b a − b l5/3 √ √ 2n/2 × 23n/2 = × x3 y3 ( √ ) √ √ a2 u − ab l5/3 u + ab − b2 l5/3 √ i.e. 2n = x3 y 3 √

√

(44)

(45)

From (iii) and (v) we get √

( √ ) (√ ) c xt + d z3 − f √ × e sp ( √ ) √ √ c xz 3 t − cf xt + d z 3 − df 2 √ 7k = e sp

√ 75/3 k7/3 × 71/3 k5/3 =

i.e.

(46)

(47)

From (ii) and (iv) we get √

i.e.

l5/3 ×

l2 =

√

l7/3 =

( ) √ a − 23n/2 y 3 b

×

( √ √ ) c y − tp d

(48)

( √ ) √ √ √ ac y − a tp − cy 2 23n/2 + 23n/2 y 3 tp (bd)

(49)

Important Note: Please See Annexure–(A) (pp. 17–21). Wherein we have established that the equivalent values for x3 , y 3 , z 3 ; sp , tp , up ; and 2n , 7k2 , l2 ; holds good through the parameters a, b; c, d; e, f . Substituting the equivalent values of 2n , 7k2 and l2 in the identity 2n = 7k2 + l2

(50)

8


and multiplying both sides by {(bde)

√ x3 y 3 sp }, we get

{ ) √ √ √ }( √ (bde) sp a2 u − ab l5/3 u + ab − b2 l5/3 }( √ ) { √ √ √ = (bd) x3 y 3 c xz 3 t − cf xt + d z 3 − df { √ }( √ ) √ √ √ + (e) x3 y 3 sp ac y − a tp − cy 2 23n/2 + 23n/2 y 3 tp

51(B)

There are four terms on L.H.S. and eight terms on R.H.S. of equation (B). Our aim is to compute the rational terms on both sides of the equation, term by term, sum up the rational terms on both sides, and equate them. In order to eliminate denominators under ‘a2 b’/‘ab2 ’/‘b3 ’ terms, ‘cd’/‘d2 ’ terms and ‘e’/‘f ’ terms let us multiply both sides of the equation by ( )3 ( )2 ( ) ul5/3 − 1 xtl7/3 − y 71/3 k5/3 z − 1

(52)

and again multiply both sides of the equation by √

2n/2 xsp tp l5/3 u(p+1)

(53)

in order to bring out some rational terms on both sides of equation (B). The rational terms are worked out below, after multiplying by the above terms, and on substituting for a, b; c, d; e and f . Note: In “a2 b”, “b3 ” and “ab2 ”, we get in the numerator the factor (√

ul5/3 − 1

)3 = =

(√

ul5/3 − 1

{√

)( ) √ ul5/3 + 1 − 2 ul5/3

( ) ( )} ul5/3 ul5/3 + 3 − 1 3 · ul5/3 + 1

(54) (55)

I{( term)in } L.H.S. of equation (B), after multiplying by the concerned terms and on substituting for a2 b de =

√

( )√ sp u xtl7/3 − y 2n/2 xsp tp l5/3 u(p+1)

)(√ ( ) √ √ × 2n/2 l5/3 x3 + 23n/2 y 3 + 2(n+1) l5/3 x3 y 3 2n/2 x3 − 23n/2 y 3 u × ×

{√

( ) ( )} ul5/3 ul5/3 + 3 − 1 3 · ul5/3 + 1

(√

) √ √ √ 75/3 k7/3 l7/3 xysp t − y 75/3 k7/3 sp − xt tp l7/3 + xyt(p+1)

( √ √ √ ) × z 2 71/3 k5/3 z − z 2 + 71/3 k5/3 zup − up

(56)

9


Rational part of this term ( )( √ ) √ √ √ = sp u 2n/2 xsp tp l5/3 u(p+1) − xt tp l7/3 − up } { ( )√ } )( )√ 2n/2 x3 (−1) − y 23n/2 y 3 2n/2 x3 (−1) xtl7/3 2n/2 l5/3 x3 {( )( )√ } )( √ √ + xtl7/3 2(n+1) l5/3 x3 y 3 − 23n/2 y 3 u ul5/3 (3) {( )( )√ ( )} + xtl7/3 23n/2 y 3 2n/2 x3 − 3ul5/3

 {(   ×  

     

( ) ( )  − 2n x7 sp t(p+2) u(p+1) l6 + 22n x3 y 4 sp t(p+1) u(p+1) l2 ) ( )  = ( − 2(2n+1) 3 · x4 y 3 sp t(p+2) u(p+2) l6 − 22n · 3 · x4 y 3 sp t(p+2) u(p+2) l6

(57)



(58)

II term of equation (B), after multiplying by the concerned terms, and on substituting {( in)L.H.S. } for ab2 de =−

(√

)( )√ (√ ) √ l5/3 sp u xtl7/3 − y 2n/2 xsp tp l5/3 u(p+1) 2n/2 l5/3 x3 + 23n/2 y 3

( ){√ ( ) ( )} √ × 2n/2 x3 + 23n/2 y 3 u − 2(n+1) x3 y 3 u ul5/3 ul5/3 + 3 − 1 3ul5/3 + 1 ×

(√ ) √ √ √ 75/3 k7/3 l7/3 xysp t − y 75/3 k7/3 sp − xt tp l7/3 + xyt(p+1)

( √ √ √ ) × z 2 71/3 k5/3 z − z 2 + 71/3 k5/3 zup − up

(59)

Rational part of this term )√ ( )( √ ) (√ √ − up = − l5/3 sp u 2n/2 xsp tp l5/3 u(p+1) − xt tp l7/3  {( ×  ( =

)√ ( ) }  xtl7/3 2n/2 l5/3 x3 2n/2 x3 + 23n/2 y 3 u (−1) {( )√ ( )√ }  √ + xtl7/3 23n/2 y 3 − 2(n+1) x3 y 3 u ul5/3 (3)

) ( )  2n x7 sp t(p+2) u(p+1) l6 + 22n x4 y 3 sp t(p+2) u(p+2) l6  ( ) + 2(2n+1) · 3x4 y 3 sp t(p+2) up+2 l6

(60)

III term of equation (B), after multiplying by the concerned terms, and on substituting {( in } ) L.H.S. for ab2 de =

)√ √ ( 7/3 −y sp xtl 2n/2 xsp tp l5/3 u(p+1) × × ×

(√

2n/2 l5/3 x3 +

{√

√

)( 23n/2 y 3

2n/2 x3 + 23n/2 y 3 u − 2(n+1)

) √ x3 y 3 u

( ) ( )} ul5/3 ul5/3 + 3 − 1 3ul5/3 + 1

(√

) √ √ √ 75/3 k7/3 l7/3 xysp t − y 75/3 k7/3 sp − xt tp l7/3 + xyt(p+1)

10


( √ √ √ ) × z 2 71/3 k5/3 z − z 2 + 71/3 k5/3 zup − up

(61)

Rational part of this term {√ ( )( √ ) √ √ = sp · 2n/2 xsp tp l5/3 u(p+1) − xt tp l7/3 − up  {(

)√ )√ } (  xtl7/3 2n/2 l5/3 x3 2n/2 x3 + 23n/2 y 3 u ul5/3 (3) ( {( )( ) }{√ )}  √ + xtl7/3 − 3ul5/3 − y(−1) 23n/2 y 3 − 2(n+1) x3 y 3 u

×  (

)( )  2n x3 + 22n y 3 u 3x4 sp t(p+2) u(p+1) l6 )( )  = ( − 2(2n+1) x3 y 3 sp t(p+1) u(p+1) l2 y − 3xtul4  ( =

) ( )  3 · 2n x7 sp t(p+2) u(p+1) l6 + 3 · 22n x4 y 3 sp t(p+2) u(p+2) l6 ( ) ( )  − 2(2n+1) x3 y 4 sp t(p+1) u(p+1) l2 + 2(2n+1) · 3x4 y 3 sp t(p+2) u(p+2) l6

(62)

IV term in { } L.H.S. of equation (B), after multiplying by the concerned terms, and on substituting 3 for b de )( )√ (√ 2n/2 xsp tp l5/3 u(p+1) = − sp l5/3 xtl7/3 − y ( )(√ ) √ √ × 2n/2 x3 + 23n/2 y 3 u − 2(n+1) x3 y 3 u 2n/2 x3 − 23n/2 y 3 u × ×

{√

( ) ( )} ul5/3 ul5/3 + 3 − 1 3 · ul5/3 + 1

(√ ) √ √ √ 75/3 k7/3 l7/3 xysp t − y 75/3 · k7/3 sp − xt tp l7/3 + xyt(p+1)

( √ √ √ ) × z 2 71/3 k5/3 z − z 2 + 71/3 k5/3 zup − up

(63)

Rational part of this term { √ ( )( √ )} √ √ = − sp · l5/3 2n/2 xsp tp l5/3 u(p+1) − xt tp l7/3 − up  {( ×

)( )√ }  √ xtl7/3 2n/2 x3 + 23n/2 y 3 u 2n/2 x3 ul5/3 (3) {( )( )( √ )√ }  √ + xtl7/3 − 2(n+1) x3 y 3 u − 23n/2 y 3 u ul5/3 (3)

( ) ( )  − 2n · 3x7 sp t(p+2) u(p+1) l6 − 22n · 3x4 y 3 sp t(p+2) u(p+2) l6  ) = ( − 2(2n+1) 3 · x4 y 3 sp t(p+2) u(p+2) l6 

(64)

Sum of all rational terms on L.H.S. of equation (B) ( )( ) = 22n x3 y 3 sp t(p+1) u(p+1) l2 y − 2xtul4 { } Combining I and II terms

(65)

11


( ) − 2(2n+1) x3 y 4 sp t(p+1) u(p+1) l2 {

} Combining III and IV terms

(66)

( )( ) = − 22n x3 y 3 sp t(p+1) u(p+1) l2 2xtul4 + y

(67)

Let us find out all rational terms on R.H.S. of equation (B). There will not any rational in the first )( ) parts√ ( be ( and ) third terms on R.H.S. of equation (B), since they √ are having x2 y 3 z 3 t bcd and x3 y 3 z 3 bd2 respectively in their leading terms, and in b, c, d √ √ and d2 we do not have any z 3 or z term for bringing out any rational terms [please see item (7) of the assumption in which we have shown that in the first term in R.H.S. √ of equation √ (B), we get 3 sp tp or some rational terms, multiplied by one of the irrational radical factors z lz 3 sp tp up or √ xyz 3 sp up . Similarly, if we multiply the following terms on III term of R.H.S. of equation (B), after multiplying by the concerned terms, then we get ( )( ) √ ( )( )( ) √ √ √ x3 y 3 z 3 − 2ul5/3 − 1 × 2n/2 xsp tp l5/3 up+1 − u 23n/2 y 3 l5/3 xtp+1 − 2 xytl7/3 √ ( ) √ ( )√ = 2n+2 x3 y 3 tp+1 u2 l2 tp+1 l5/3 l5/3 (xyz 3 )sp up+1 {( =

2n+2 x3 y 3 tp+1 u2

)√

(tu)p+1 l4

√

} l(xyz 3 )sp

(68)

which is irrational, since x, y, z 3 or square-free integer, each pair-wise relatively prime to l and s II term of equation (B), after multiplying by the concerned terms, and on substituting { in R.H.S. } for b(cd)f ( √ )( )2 √ = − x2 y 3 t ul5/3 − 1 2n/2 xsp tp l5/3 u(p+1) ×

(√

2n/2 x3 ul5/3 −

√

2n/2 x3 − u

) √ √ 23n/2 y 3 l5/3 + 23n/2 y 3 u

( √ √ √ √ ) × 75/3 k7/3 sp l7/3 y − 75/3 k7/3 l7/3 xsp t(p+1) + 75/3 · k7/3 sp tp y − tp xt ( ) √ × xtl7/3 + y − 2 xytl7/3 ( √ ) √ √ √ × z 2 71/3 · k5/3 − z 3 − 71/3 k5/3 zup + 71/3 k5/3 up

(69)

Rational part of this term [ ( √ ) √ √ = − x2 y 3 t (1) 2n/2 xsp tp l5/3 u(p+1) 23n/2 y 3 u ×

{

−

√

}{ 75/3 k7/3 l7/3 xsp t(p+1)

y + 2y

}√

( )( ) = 3 · 2n · 7 · k2 · l2 x3 y 4 sp t(p+1) u(p+1)

] 71/3 k5/3 up

(70) (71)

12


IV term in { } R.H.S. of equation (B), after multiplying by the concerned terms, and on substituting for bd2 f =−

(√

×

)( x3 y 3

(√

)2 √ ul5/3 − 1 2n/2 xsp tp l5/3 u(p+1)

2n/2 x3 ul5/3 −

√

) √ √ 2n/2 x3 − u 23n/2 y 3 l5/3 + 23n/2 y 3 u

( ) √ × 75/3 k7/3 sp y + xtp+1 − 2 75/3 · k 7/3 xysp t(p+1) ( ) √ × xtl7/3 + y − 2 xytl7/3 ( √ ) √ √ √ × z 2 71/3 · k5/3 − z 3 − 71/3 k 5/3 zup + 71/3 k5/3 up

(72)

Rational part of this term [ (√ ) √ √ = − x3 y 3 (1) 2n/2 xsp tp l5/3 u(p+1) 23n/2 y 3 u ×

(

−2

)( )√ ] √ √ 75/3 · k7/3 xysp t(p+1) − 2 xytl7/3 71/3 k5/3 up

( )( ) = − 4 · 2n · 7 · k 2 l2 x3 y 4 sp tp+1 up+1

(73) (74)

V term { in } R.H.S. of equation (B), after multiplying by the concerned terms, and on substituting for ace ( √ )( )2 ( )√ = y 2 x3 sp ul5/3 − 1 xtl7/3 − y 2n/2 xsp tp l5/3 u(p+1) ) ( √ √ √ √ × l5/3 2n/2 x3 u − 2n/2 l5/3 x3 + 23n/2 y 3 ul5/3 − 23n/2 y 3 ( √ √ √ √ ) × l7/3 75/3 k7/3 xsp t − 75/3 · k7/3 l7/3 sp y + xt(p+1) l7/3 − ytp ( √ √ √ ) × z 2 71/3 k5/3 z − z 2 + 71/3 k 5/3 zup − up Rational part of this term √ √ √ √ √ = y 2 x3 sp {1 · xtl7/3 + u2 l10/3 (−y)} 2n/2 xsp tp l5/3 u(p+1) 23n/2 y 3 ul5/3 (− ytp )(− up ) =

[( )( ) ] [ ] 2n · x3 y 4 sp t(p+1) u(p+1) l4 − 2n x2 y 5 sp tp+1 up+3 l5

(75)

(76) (77)

VI term { } in R.H.S. of equation (B), after multiplying by the concerned terms, and on substituting for ae =−

(√

)( x3 y 3 sp tp

ul5/3 − 1

)2 ( )2 √ xtl7/3 − y 2n/2 xsp tp l5/3 u(p+1)

( ) √ √ √ √ × l5/3 2n/2 x3 u − 2n/2 l5/3 x3 + 23n/2 y 3 ul5/3 − 23n/2 y 3 ( √ √ √ ) × z 2 71/3 k 5/3 z − z 2 + 71/3 k5/3 zup − up

(78)

13


Rational part of this term ( √ ) √ √ √ = − x3 y 3 sp tp {1 · (−2xytl7/3 ) + u2 l10/3 y 2 } 2n/2 xsp tp l5/3 u(p+1) 23n/2 y 3 ul5/3 − up (79) ( )( ) ( ) = − 2(n+1) x3 y 4 l4 sp t(p+1) u(p+1) + 2n x2 y 5 l5 sp tp+1 up+3

(80)

VII {term } in R.H.S. of equation (B), after multiplying by the concerned terms, and on substituting for

ce )( )3 ( )√ ( √ xtl7/3 − y 2n/2 xsp tp l5/3 u(p+1) = − y 2 23n/2 x3 y 3 sp ul5/3 − 1 ( √ √ √ √ ) × l7/3 75/3 k7/3 xsp t − 75/3 · k7/3 l7/3 sp y + xtp+1 l7/3 − ytp ( √ √ √ ) × z 2 71/3 k5/3 z − z 2 + 71/3 k5/3 zup − up

(81)

There is no rational part in this term. Note: If we multiply { ( √ ) (√ )( √ ))} √ − y 2 23n/2 x3 y 3 sp (−1)(−y) 2n/2 xsp tp l5/3 up+1 xtp+1 l7/3 − up (82) ( )√ then we get 2n x2 y 4 sp tp up l2 xytu, which is irrational since x and y are square-free integers, gcd(x, y) = 1, and both are relatively prime to t and u. [See Annexure–(B) (pp. 22–24)]. VIII term in R.H.S. of equation (B), after multiplying by the concerned terms, and on substituting for ‘e’ ( √ )( )3 ( )2 √ = y 3 23n/2 x3 sp tp ul5/3 − 1 xtl7/3 − y 2n/2 xsp tp l5/3 u(p+1) ( √ √ √ ) × z 2 71/3 k5/3 z − z 2 + 71/3 k5/3 zup − up (83) There is no rational part in this term. ( ) ( )( ) ( ) ul5/3 − 13 = ul5/3 − 1 u2 l10/3 + 1 − 2ul5/3 = u3 l5 + 3u15/3 − 3u2 l10/3 − 1

(84)

Note: If we multiply the terms {( √ )( ) √ ( √ )} y 3 23n/2 x3 sp tp 3ul5/3 y 2 2n/2 xsp tp l5/3 u(p+1) − up , (85) ( ) √ then we get − 3 · 2n x2 y 5 sp tp up+1 l2 lu, which is irrational since we have chosen l as a square-free integer, relatively prime to u, as discussed earlier. [See Annexure–(B) (pp. 22–24)]. Sum of all rational terms on R.H.S. of equation (B) ( )( )( ) { } = − 2n · 7 x3 y 4 k2 l2 sp t(p+1) u(p+1) Combining II and IV terms ( )( ) { } − 2n · x3 y 4 l4 sp t(p+1) u(p+1) Combining V & VI terms ( )( )( ) = − 2n · x3 y 4 l2 sp t(p+1) u(p+1) 7k2 + l2 ( )( ) [ ( ) ] = − 22n · x3 y 4 l2 sp t(p+1) up+1 ∵ 7k2 + l2 = 2n

(86) (87) (88)

14


Equating the rational terms on both sides of equation (B), we get ( )( ) ( )( ) − 22n x3 y 3 sp t(p+1) u(p+1) l2 2xtul4 + y = − 22n x3 y 4 l2 sp t(p+1) up+1 ( ) Dividing both sides by − 22n x3 y 3 l2 , we get ( )( ) ( ) sp t(p+1) u(p+1) 2xtul4 + y = y sp tp+1 up+1 ( )( ) sp t(p+1) u(p+1) 2xtul4 = 0 ( ) Dividing both sides by 2xl4 we get ( ) sp tp+2 up+2 = 0 i.e.

(89)

(90) (91)

(92)

i.e. Either s = 0 or t = 0 or u = 0. This contradicts our hypothesis that all s, t and u are non-zero positive integers in the Fermat’s equation sp + tp = up , p a prime > 3 (93) and shows only a trivial solution is possible with stu = 0. This is the end of the proof.

4

Conclusion

Since equation (B) in this proof was obtained from the equivalent equations to Fermat’s equations x3 + y 3 = z 3 and

sp + tp = up

the result we get by solving equation (B) must necessarily reflect on the original Fermat’s equations. The only hypothesis we made in this proof is that all s, t and u are natural numbers, which is shattered by the result stu = 0, showing that only a trivial solution is possible in the Fermat’s equation sp + tp = up .

Competing Interests Author has declared that no competing interests exist.

References [1] Yves Hellegouarch. Invitation to the mathematics of Fermat. Wiles. 2002;1-31. [2] Richard A. Mollin, algebraic number theory. (Chapman & Hall/CRC). Page 59: Theorem: Solutions to Ramanujan - Nagell equation. 2011;4-5(59):64-127. [3] David M. Burton tata, elementary number Theory VI Edition. McGraw-Hill Edition. 2006;256-257.

15


[4] Hardy GH, Wright EM. An introduction to the theory of numbers VI Edition. Oxford University Press. 2008;261:586. [5] Joseph H. Silverman jorn tate, rational points on elliptic Curves. Undergraduate Texts in Mathematics. Springer. 1992;230. [6] Gabor Toth. Glimpses of algebra and geometry ii edition. Springer. 2002;50-52. [7] Abc Conjecture (also known as the Oesterle–Masser conjecture) (From Wikipedia). 2017;1-20. [8] Nicolas Billerey, Imin Chen, Lusi Dieulefait, Nuno Freitas. A result on the equation xp + y p = z r using Frey abelian varieties. Proceedings of the American Mathematical Society. [Published Electronically: June 16, 2017]. 2017;1-24. [9] Andrew Wiles. Modular elliptic curves and Fermat’s Last Theorem. Annals of Mathematics. 1995;141(3):443-551. [10] Nigel Boston. The proof of fermat’s last Theorem. University of Wisconsin–Madison Spring; 2003. [Introduction and Chapters 1 to 12]. 2003;1-140. [11] Problem solving strategies Arthur Engel Edited by K. Benscath, P. R. Halmos Springer. [Chapter 6: Numbery Theory Item E14]. 1998;126-127.

16


Annexure–A We can easily show that the equivalent values of x3 , y 3 , z 3 ; sp , tp , up ; and 2n , 7k2 and l2 holds good through parameters a, b; c, d; e and f as worked out below: 1) From the equivalent equations we have √ √ a u+b x3 = √ (1) 2n/2 where

(√ a=

and

) √ 2n/2 l5/3 x3 + 23n/2 y 3 (√ ) l5/3 u + 1

(√ b=

) √ 2n/2 x3 − 23n/2 y 3 u ) (√ l5/3 u + 1

(2)

(3)

Substituting the above values of a and b in the above identity we get {√ (√ ) √ } √ √ u 2n/2 l5/3 x3 + 23n/2 y 3 + 2n/2 x3 − 23n/2 y 3 u √ (√ ) x3 = √ 2n/2 l5/3 u + 1 {√ =

(√ )(√ )} x3 2n/2 l5/3 u + 1 √ (√ ) = x3 . √ 2n/2 l5/3 u + 1

2) From the equivalent equations we have √ y3 =

(

) √ a − b l5/3 √ 23n/2

(4)

(5)

Substituting the values of a and b in the above identity, we have {(√ ) √ (√ )} √ √ 2n/2 l5/3 x3 + 23n/2 y 3 − l5/3 2n/2 x3 − 23n/2 y 3 u √ (√ ) y3 = √ 23n/2 l5/3 u + 1 = =

√ √

( )/(√ )√ √ 23n/2 y 3 1 + l5/3 u l5/3 u + 1 23n/2 y3 .

(6)

3) From the equivalent equations we have √ where

(√ c=

sp

√ c xt + d = √ 75/3 k7/3

√ ) 75/3 k7/3 l7/3 sp + tp (√ √ ) xtl7/3 + y

(7)

(8)

17


and

(√ d=

) √ 75/3 k7/3 sp y − xtp+1 √ √ xtl7/3 + y

(9)

Substituting the above values of c and d in the above identity we have, )} {√ (√ √ √ ) (√ xt 75/3 k7/3 l7/3 sp + tp + 75/3 k 7/3 sp y − xtp+1 √ (√ sp = √ √ ) 75/3 k7/3 xtl7/3 + y √ =

=

√

√ (√ √ ) 75/3 k7/3 sp xtl7/3 + y (√ √ √ ) 75/3 k7/3 xtl7/3 + y

sp .

(10)

4) From the equivalent equations, we have √ √ √ tp = c y − d l7/3

(11)

where c and d are having values as given in item (3) above. Substituting the above values of c and d in the above identity we have (√ )} {√ (√ √ √ ) √ y 75/3 k7/3 l7/3 sp + tp − l7/3 75/3 k7/3 sp y − xtp+1 √ (√ tp = √ ) xtl7/3 + y =

{√ √ }/(√ √ √ √ ) y tp + tp xtl7/3 xtl7/3 + y

=

)/(√ √ √ (√ √ ) √ tp y + xtl7/3 xtl7/3 + y = tp .

5) From the equivalent equations, we have ( √ ) √ z 3 = e 71/3 k5/3 + f where

( ) √ )/(√ e = z 2 + up 71/3 k5/3 z + 1

and f=

(√ )/(√ ) √ z 3 − 71/3 k5/3 up 71/3 k5/3 z + 1

(12)

(13)

(14)

(15)

Substituting the above values of e and f in the above identity, we have ( )} {√ √ √ ) (√ 71/3 k5/3 z 2 + up + z 3 − 71/3 k5/3 up √ ) (√ z3 = 71/3 k5/3 + 1 ( √ ) √ )/(√ = z 2 71/3 k5/3 + z 3 71/3 k5/3 z + 1 = =

)/(√ ) √ (√ z3 71/3 k5/3 z + 1 71/3 k5/3 z + 1 √

z3.

(16)

18


6) From the two equivalent equations, we get √ √ up = e − f z

(17)

Substituting the above values of e and f , we have ( ) √ √ ) √ (√ z 2 + up − z z 3 − 71/3 k5/3 up √ ) (√ up = 71/3 k5/3 z + 1 = =

√ √

( )/(√ ) √ up 1 + 71/3 k5/3 z 71/3 k 5/3 z + 1 up .

(18)

7) From the two equivalent equations, we get ( √ ) √ √ a2 u − ab l5/3 u + ab − b2 l5/3 u √ 2n = x3 y 3

(19)

where (√ a=

) √ 2n/2 l5/3 x3 + 23n/2 y 3 (√ ) l5/3 u + 1

(20)

and (√ b=

) √ 2n/2 x3 − 23n/2 y 3 u (√ ) l5/3 u + 1

(21)

Substituting the √ above√values of a and d in the above identity, we get, on multiplying both sides by x3 y 3 ( l5/3 u + 1)2 , 2n

√

( (1) √(3) ) (2) x3 y 3 l5/3 u + 1 + 2 l5/3 u

=

(5) ) √(3) √ ( n/2(4) u 2 l5/3 x3 + 23n/2 y 3 + 2n+1 l5/3 x3 y 3

−

√ (

+

−

(4) (6) (7) ( √ (1) √ √ √ ) l5/3 u 2n/2 x3 l5/3 − 2n l5/3 x3 y 3 u + 2n x3 y 3 − 23n/2 y 3 u

(8) (2) (5) √ (6) √ √ √ ) 2n/2 x3 l5/3 − 2n l5/3 x3 y 3 u + 2n x3 y 3 − 23n/2 y 3 u u

√

(6) (7) ( (8) ) √ l5/3 2n/2 x3 + 23n/2 y 3 u − 2n+1 x3 y 3 u

(22)

we find that all terms getting cancelled on both sides; the corresponding terms getting cancelled out are marked as (1), (2), (3), (4), (5), (6) (at 3 places), (7) and (8) for easy identification.

19


8) From the two equivalent equations we get (

7k2

)( √ ) {( √ (√ )} √ ) e sp = c xz 3 t − cf xt + d z 3 − f

(23)

where (√ c=

√ ) 75/3 k7/3 l7/3 sp + tp (√ ; √ ) xtl7/3 + y

( e = (√

z2 +

√

) up

71/3 k5/3 z + 1

);

(√ d=

) √ 75/3 k7/3 sp y − xtp+1 (√ √ ) xtl7/3 + y

(√ ) √ z 3 − 71/3 k5/3 up ) f = (√ 71/3 k5/3 z + 1

(24)

(25)

Substituting the of c, d, e and f in the above identity after multiplying √ above values √ √ both sides by ( xtl7/3 + y) ( 71/3 k5/3 z + 1) we get ( √ (3) ( ){( √ )( √ (1) (2) ) (4) )} √ √ √ xtl7/3 + y + sp up xtl7/3 + y 7k2 z 2 sp

=

{√

( √ (1) ( √ (6) √ (7) ) (5) )} xz 3 75/3 k7/3 l7/3 sp t 71/3 k5/3 z + 1 + xz 3 tp+1 71/3 k5/3 z + 1

−

(7) (8) ) ( √(5) )} √(3) √ √ {( √ xt 75/3 k7/3 l7/3 sp z 3 − 7 · k2 l7/3 sp up + z 3 tp − 71/3 k 5/3 tp up

+

(9) (6) (2) {√ ( √ √ √ (10) )} √ z 3 7k2 sp yz + 75/3 k 7/3 sp y − xz · 71/3 k5/3 tp+1 − xtp+1

−

(9) (8) (4) (√ ) √ √ (10) √ 75/3 · k7/3 sp yz 3 − 7 · k2 sp yup − xz 3 tp+1 + xtp+1 up 71/3 k5/3

(26)

All terms getting cancelled on both sides. The corresponding terms get cancelled are marked as (1), (2), (3), (4), (5), (6), (7), (8), (9) and (10) for easy identification. 9) Finally from the two equivalent equations, we get ) ( ) ( √ √ √ √ l2 bd = ac y − a tp − cy 2 23n/2 + 23n/2 y 3 tp

(27)

where (√ a= (√ c=

) √ 2n/2 l5/3 x3 + 23n/2 y 3 (√ ) ; l5/3 u + 1 √ ) 75/3 k7/3 l7/3 sp + tp (√ √ ) xtl7/3 + y

(√ b=

) √ 2n/2 x3 − 23n/2 y 3 u (√ ) l5/3 u + 1

(√ and

d=

) √ 75/3 k7/3 sp y − xt(p+1) (√ √ ) xtl7/3 + y

(28)

(29)

20


Substituting the above values above equivalent equation, after √ of a, b, c and √ d in the √ multiplying both sides by ( l5/3 u + 1) ( xtl7/3 + y) we get l2

(4) (1) (3) {√ } √ √ (2) √ 2n/2 · 75/3 k7/3 sp x3 y − x2 2n/2 tp+1 − y 2 23n/2 · 75/3 k7/3 sp u + 23n/2 xy 3 tp+1 u

=

(6) (7) (1) (5) √ { 2 √ n/2 5/3 7/3 3 p √ n/2 5/3 3 p √ 3n/2 5/3 7/3 7/3 3 p √ 3n/2 3 p } y l 2 ·7 k x s + 2 l x t + 2 ·7 k l y s + 2 y t

−

(5) (8) (2) ) √ √ √(9) √ ( 2 2√ tp x l 2n/2 t + 2n/2 l5/3 x3 y + 23n/2 xy 3 tl7/3 + y 2 23n/2

(6) (7) ) ( √ (3) √ √ √ (10) √ − y 2 23n/2 l2 75/3 k7/3 sp u + 75/3 k7/3 l7/3 sp + l5/3 tp u + tp

+

(4) ( √ (9) ) √ √ (10) √ (8) √ 23n/2 y 3 tp l2 xtu + l5/3 yu + xtl7/3 + y

(30)

All terms are getting cancelled. The corresponding terms get cancelled are marked as (1), (2), (3), (4), (5), (6), (7), (8), (9) and (10) for easy identification.

21


Annexure–B In the computation of rational terms in Equation (B) in the main proof after multiplying both sides of the equation by [( )3 ( )2 ( )√ ] ul5/3 − 1 xtl7/3 − y 71/3 k5/3 z − 1 2n/2 xsp tp l5/3 u(p+1) (31) we are getting three types of rational terms i) Rational terms which do not have any radical factor. ii) Rational terms multiplied by any one of the radical factors ( ) √ √ z 3 sp tp ; xyz 3 sp l; t1/2 2n/2 x3 + 23n/2 y 3 u ; √ √ √ √ xyz 3 sp up ; xytu; lz 3 sp tp up ; yz 3 sp up

(32) (33)

Since we consider n as an odd integer and also t as an odd integer, (in the equations 2n = 7k2 + l2 and sp + tp = up ), (t1/2 )(2n/2 x3 + 23n/2 y 3 u) will be irrational. Also since we have chosen x, y, z 3 as square-free integers; gcd(x, y, z 3 ) = 1 and √ each 3 z 3 sp tp ; of x, y, z is pair-wise relatively prime to s, t, u and l, all the radical terms √ √ √ √ √ xyz 3 sp l; xyz 3 sp up ; xytu; lz 3 sp tp up and yz 3 sp up (as worked out in the main proof), in which x, y, z 3 , in some combinations with s, t, u and l, √ are inside a square-root, will be irrational. This is explained with one example: Suppose xyz 3 sp up is an integer equal to I (say), then I 2 = xyz 3 sp up . In this we find all the prime factor of I appear as square numbers in L.H.S., whereas on the right side we have x, y and z 3 as square-free 3 3 integers; gcd(x, y, √z ) = 1 and each of x, y, z is chosen as pair-wise relatively prime to s and u. Hence xyz 3 sp up is irrational. √ √ √ √ √ Similarly we can prove that xyz 3 sp l, xytu, z 3 sp tp , lz 3 sp tp up and yz 3 sp up are irrational radicals. In order to avoid such radical factor appearing in the computation of rational terms in Equation (B) under the main proof and for the sake of convenience we have chosen x, y and z 3 as square-free integers, with gcd(x, y, z 3 ) = 1 and all x, y and z 3 are pair-wise relatively prime to s, t, u and all l. Since our aim is to prove Fermat’s last theorem in the equation s p + tp = u p ,

(34)

p a prime > 3 with the hypothesis that there exist some integers (non-zero positive) s, t and u, and bringing contradiction (proving by method of contradiction) we are at liberty in choosing x, y and z 3 in the Fermat’s equation x3 + y 3 = z 3

(35)

as non-zero positive integers, z and z 2 irrational, we have used the proof of FLT proved for n = 3, and using this proof we try to prove the theorem for n = p, where p is any prime > 3. iii) We are getting some rational terms in the first four terms on L.H.S. and in the second, fourth, fifth, sixth, seventh and eighth terms on R.H.S. of√ equation (B)in the main proof, 5/3 in which we get some rational terms multiplied by (l l5/3 u), that is, some rational √ terms multiplied by (l2 lu). Such terms have not been worked out completely in all terms in equation (B), and only √ some examples have been worked out, since we can consider that lu is irrational, due to the following reasons:

22


1) l and u are connected by the Euler’s and Fermat’s equations n ≥ 3;

2n = 7k2 + l2 ;

k, l

odd

(36)

and s p + tp = u p ;

p a prime > 3

(37)

we have chosen n as an odd integer and t as an odd integer. Clearly we see k and l as odd integers. We have chosen k and l as > 1, neither of them perfect squares nor perfect cubes. 2) If n ≥ 3, then 2n can be represented in the form 2n = 7k2 + l2

with odd

k, l.

(38)

Figure below shows the first 8 solutions which can easily be found by guessing. Now study this table closely. Before reading on, try to find the pattern behind the table

n k l

3 1 1

4 1 3

5 1 5

6 3 1

7 1 11

8 5 9

9 7 13

10 3 31

Our hypothesis is that one column somehow determines the next one. How I can get the next pair k1 , l1 form the current k, l? This conjecture is supported by similar equations, for instance the Pell-Fermat equation where we get form one pair (k, l) to the next by a linear transformation. Let us start with k1 . How can I get from (k, l) to k1 ? We get k1 from the first pair (1, 1) by the arithmetic mean. From the second pair (1, 3), the mean 2 is not an odd integer. So let us take the difference |k − l|/2 = 1. Again we are successful. Some more trials convince us that we should take (k + l)/2 if that number is odd. If that number is even we should take |k − l|/2. After guessing the pattern behind k, we will try to guess the pattern behind l. There is a 7 before k2 in the equation. So we could try (7k + y)/2 and |7k − y|/2. The pattern seems to hold for the table above. To support our conjecture, we observe that exactly one of k+l 2

or

|k − l| 2

or

|7k − l| 2

is odd

(39)

|k − l| k+l + = max(k, l). 2 2 Exactly one of since

7k + l 2 since

is odd

(40)

|7k − 1| 7k + l + = max(7k, l). 2 2

23


In addition, we have k+l 2

|k − l| 2

odd ⇒

|7k − l| 8k − (k + l) = 2 2 k + l = 4k − 2

odd ⇒

odd

(41)

8k − (k − l) 7k + l = 2 2 = 4k −

k−l 2

odd.

So we have the following transformations: ( ) k + l |7k − l| S : (k, l) 7→ , , 2 2 ( ) |k − l| 7k + l T : (k, l) 7→ , 2 2

(42)

(43)

(44)

Now we prove our conjecture by induction. It is valid for n = 3. Suppose 7k 2 + l2 = 2n for any n. By applying s, we get 7(k + l)2 + (7k − l)2 = 14k2 + 2l2 4 = 2(7k 2 + l2 ) = 2 · 2n = 2n+1 .

(45)

Similarly we can proceed with transformation T . [Courtesy: Problem solving strategies ARTHUR ENGEL see: Bibliography item [2]]. Therefore we may derive infinitely many values of l as an odd positive integer from the Euler’s equation 2n = 7k2 + l2

(46)

wherein we have considered n as any odd integer ≥ 3, for convenience, in this proof. Hence depending upon the value of n as an odd integer, we will get l as an odd integer for infinitely many values such that l is square-free and relatively prime to u and therefore √ lu can be considered irrational. ——————————————————————————————————————————————c 2017 Seetharaman; This is an Open Access article distributed under the terms of the Creative ⃝ Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Peer-review history: The peer review history for this paper can be accessed here (Please copy paste the total link in your browser address bar) http://sciencedomain.org/review-history/21048

24