Exciting multiple proofs of FLT  

M.Meyyappan, Professor (Retd), Department of Physics, Alagappa Government Arts College, Karaikudi-630003,Tamilnadu,India meydhanam@gmail.com 

Abstract 

The Diophantine equation an+bn=cn, where a, b, c, and n are integers, has no nonzero integer solutions for n>2 is called Fermat's Last Theorem. Since its prediction in the 16th century many mathematicians tried to proof FLT for more than 350 years, However .It came to an end in 1995, with a successful proof of FLT by Taylor and Wills .This paper describes some of the simplest and exciting proofs of FLT 

Key words 

Number theory-Fermat’s Last Theorem - Beal’s conjecture – equal power relation 

1. Introduction 

In number theory FLT states that no three positive integers a,b,c satisfying the equation an + bn = cn for all exponents n ≥ 3. If a,b,c are whole integers their exponents will not be same and for the same exponents a.b.c will not be whole integers. After the Fermat’s assertion many people [1.2.3. ] tried to prove FLT for different values of exponents using number theory. The computer supported the verification of FLT for higher exponents from 3 to 4,000,000 [4] A general proof is given by Wills [5,6] with the help of elliptic curve and modular theory. However there are many different ways to prove FLT with known arithmetic axioms and the fundamentals of Number theory. The equality is determined by the same degree of evenness/oddness, same remainder for a given divisor,   same number-digit and same derivatives , if the equal power relation is in algebraic form. By solving the algebraic relations derived for an equal or multi-power relations, one can prove FLT from the non-existence of integral solution . FLT can also be proved by showing that any one of the members in the sum of three powers equal to same power of a number [an = bn + cn + dn] and in the sum of two likepowers equal to sum of same power of two other numbers [ an + bn = cn + dn] cannot be reduced to zero without disturbing the balance. 

2. Proof-1: Prime factor method [7.8] 

Any number can be represented in terms of its prime factors. An odd number will have only odd prime factors denoted by Po, where as all even numbers will invariably have 2 as one of the prime factors and besides that it may have one or more odd prime factors. 

N(odd) = P01α P β P γ......................... P ω N(even) = 2m P α P β ........................... P ω  

Depending the value of a,b,c ,   α,β,γ,...........ω may have any value including 0 . m has a minimum value of 1 in the case of singly even and its value increases depending upon the degree of evenness. If all the powers are zero, the base number becomes 1 and for a number greater than unity, at least one or more exponents will be non-zero. FLT can be proved with the knowledge of prime factors associated with one of the members of the triple (a,b,c)..

01       02       03

In aⁿ + bⁿ = cⁿ, aⁿ is equal to a product of two factors (c^n/2 - b^n/2) (c^n/2 + b^n/2) and if the prime factors associated with a (or b) is known , then they can be distributed for these two factors. By solving the two equations, one can determine the unknown base numbers in terms of the known. For simplicity let us assume a = P ^α P ^β P ^γ (or assume b )

01         02         03

an  = P nα P nβ P nγ = (cn/2 - bn/2) (cn/2 + bn/2)

In any three member equal power relation aⁿ + bⁿ = cⁿ, c is always greater than both a and b, but c^m is less than a^m + b^m when m < n and greater when m > n . Since (c^n/2 - b^n/2) < a^n/2,

02            03

01             02            03

cn/2 + bn/2 = k[P   nα/2 P   nβ/2 P   nγ/2] = k an/2 cn/2 - bn/2  = [P01nα/2 P nβ/2 P nγ/2]/k = an/.2 / k

where k >1 , need not be an integer and may have all possible values including fraction, complex, and irrational.

Solving these two equations, we get,

 

01         02         03

01         02         03

cⁿ = P ^nα P ^nβ P ^nγ [(k² + 1)/2k]² = aⁿ [(k² + 1)/2k]² bⁿ = P ^nα P ^nβ P ^nγ [(k² – 1)/2k]² = aⁿ [(k² – 1)/2k]²

and substituting these values , we get

 

aⁿ + aⁿ [(k² - 1)/2k]² = aⁿ [(k² + 1)/2k]² ≡ aⁿ + bⁿ = cⁿ

If one is able to convert it into aⁿ + bⁿ = cⁿ with a,b,c are all positive integers, it is equal to say that FLT is disproved. It is noted that all such relations irrespective of exponents n ≥ 3.the components (a,b,c) have a common factor . To make the components b and c to be whole numbers,

Cⁿ = aⁿ [(k² + 1)/2k]² = aⁿ γⁿ = (aγ)ⁿ bⁿ = aⁿ [(k² - 1)/2k]² = aⁿ βn = (aβ)ⁿ

 

                                                                      -4-

where β, γ are related together by γⁿ - βn = 1. This essential requirement gives a valid proof for FLT as no whole numbers or fractions satisfy this condition except for n = 2.

     In [a,b=c]ⁿ a,b,and c are directly proportional to 2k, (k² – 1) and (k² + 1) respectively when n = 2,but aⁿ, bⁿ and cⁿ are directly proportional to (2k)², (k² – 1)² and (k² + 1)² respectively, when n ≥ 3,where all the three in (a.b.c) cannot be made to represent positive integers for any given value of k.

The addition of γⁿ and βn gives γⁿ + βn = (k⁴ + 1)/2k². Multiplying with γⁿ - βn

= 1 , we have γⁿ + βn = γ ²ⁿ – β ²n and γⁿ (γⁿ -1) = βn (βn -1). It substantiates FLT as no set of whole numbers satisfy this condition. This can be proved by squaring γⁿ - βn = 1,

(γⁿ - βn)² = γ ²ⁿ + β ²n – 2 γⁿ βn =1

γ ²ⁿ + β ²n = 1 + 2 γⁿ βn

By adding or subtracting with γ ²ⁿ – β ²n = γⁿ + βn we have 2 γ ²ⁿ = (γ ^n/2 + β n/2)² + 1 2 β ²n = 1 - (γ ^n/2 - β n/2)²

These relations cannot be true due to non-conservation of evenness which gives an additional support to FLT.

1.  Proof-2: Algebraic method- Forbidden Fermat and allowed Beal

FLT can be proved in a simplest way with the help of algebra by taking any well balanced numeral relation a+b = (a+b) = c , where a and b have only whole integral values. By multiplying ‘a’ with a^n-1 (or ‘b’ with b^n-1) convert the assumed simple relation into a power relation aⁿ + b a^n-1 = (a+b) a^n-1 Whatever may be the value of b (e.g., b = a, aβn …… or (a+b) = a γⁿ ) not both the remaining two members b and (a+b) can have simultaneously whole integral values.

With the help of algebra one can prove FLT by showing that very same mathematical approach allows three member multi-power relations (Beal’s relations) aⁿ + bⁿ = c^m but forbids equal power relations, aⁿ + bⁿ ≠ cⁿ, where a,b,c,m,n are all integers.

The relation aⁿ + bⁿ = c^m is rewritten as (a^n/3)³ + (b^n/3)³ =   p³ + q³ = (p+q)(p² – pq + q²) = c^m. The permissible values of a.b,c for various possible sets of m,n

                                                             -5-

show that m may be greater or smaller than n but not m=n. if m = n , any two members become complex or irrational. When m ≥ n, p+q = kc and p² – pq + q²

=c^m-1/k. Solving for q (or p) , we get q = (kc/2) ±(kc/6)√[12 c^m-3/k³ – 3]. The whole integer value of q requires that the term within the square root must be a perfect square (S2) and 12 c^m-3 =k3 (S²+3) . If S = 0. p = q = kc/2. The permissible values of m and n for integral values of a,b and c are given in Table.1 . It is noted that no numeral relation is established for m = n.

Table,1, aⁿ + bⁿ = p³ + q³ = (p+q)(p² – pq + q²) = c^m.

m ≥ n                          m ≤ n

S=0, 4c^m-3 = k³

If m = 1,   c = 16, k = ¼ .                                                       2³ + 2³ = 16¹

if m = 2,   ,c= 4, k = 1                                                          2³ + 2³ = 4²

if m = 3, k³ = 4, k and hence a, b are complex or irrational or fraction, m ≠ nif m = 4, c = 16 ,k= 4.  32³ + 32³ = 16⁴

if m = 5 c = 4, k =4 ,                      8³ + 8³ = 4⁵

If m = 3α , (α= 1,2,3…..), k³ = 4 c^{3(α -1)} . k and hence a, b are complex or

,             irrational or fraction, m ≠ n S=1, 3 c^m-3 = k³

If m =1 , c = 9, k =1/3                                                       1³ + 2³ = 9¹

If m = 2 c=3, k=1                                                                1³ + 2³ = 3²

If m = 3 k³ = 3, k and hence a, b are complex or irrational or fraction, m ≠ n If m =4 ,c = 9, k 3                                                                9³ + 18³ = 9⁴

If m =5, c = 3 , k = 3                            3³ + 6³ = 3⁵

One can prove FLT from the three member multi –power relations aⁿ

± bⁿ = c^m by showing that m,n can take all possible values except m = n or one exponent is a multiple of other. We know the sum of aⁿ + aⁿ = 2a^n. To represent 2aⁿ as a power of a number without affecting the equality of the power relation, the only possibility is ‘a’ itself must be equal to 2 or 2^k. Under this condition the two different exponents can never be same . This is true in the case of (aⁿ + bⁿ). In the sum or difference of aⁿ ± bⁿ = (aⁿ ± bⁿ) . a common multiplier which makes all the terms to be a perfect power is (aⁿ ± bⁿ)ⁿ.

 

This method of getting multi-power relation provides yet another simplest proof for FLT. Sum of two identical cubes can be shown as a square or fourth or fifth power of a number but never as a single cube or with exponents, a multiple of 3.For example (2x²)³ + (2x²)³ = (2²x³)² ; (2x⁴)³ + (2x⁴)³ =

 

                                                     -6-

                                                  

(2x²)⁴ ; (2³x⁵)³ + (2³x⁵)³ = (2² x³)⁵ but (2^p x^q)³ + (2^p x^q)³ = 2^3p+1 x^3q ≠ (2^r x^s)³ = 2^3r x^3s which demands q = s and 3p+1 = 3r. If p (or r) is an integer, r (or p) cannot be.

In aⁿ + bⁿ = c^m, either aⁿ or bⁿ can be expressed as a power of m but not both proves FLT.The condition required to make n=m in aⁿ will not satisfy the additional condition required to make m= n in b^n. The non-existence of conditions satisfying both the changes resists the formation of equal power relations

All relations aⁿ + bⁿ = c^m must satisfy 1 + βⁿ/aⁿ = a^m-n or a^m-n γ^m to generate irreducible and 1 + βⁿ = a^m-n or a^m-n γ^m to generate irreducible expressions, where β and γ are integers. Whether it is reducible or irreducible m ≠ n. If m = n, an insoluble condition like 1 + βⁿ = γ^m disallows such three member equal power relation. All multi-power relations gives a practicably soluble conditions. Like equal power relation ,the three member multi-power relations have the common structure aⁿ + aⁿ [(k² – 1)/2k]² = aⁿ [(k² +1)/2k]² à^, aⁿ + aⁿ q= aⁿ r ^,where a and k are two independent variables. This can be converted into a multi-power relation by properly fixing the values of q and r. The conditions required to construct various multi-power relations are given in Table.2,which clearly show the insoluble condition in the case of three member equal power relation results with FLT.

 

Table.2. Forbidden three member equal power and allowed multi-power relation

----------------------------------------------------------------------------------------------------------

a^x + b^y = c^z                                  condition                            numerical example

 

aⁿ + bⁿ = cⁿ                                    1 + βn = γn                                        not possible (FLT)

a³ + b³ = c²                                    1 + β3 = γ2/a                                 9³ + 18³ = 81²

a³ + b³ = c⁴                                    1 + β3 = aγ4                                             9³ + 18³ = 9⁴

a³ + b³ = c⁵                                    1 + β3 = a² γ5                                          3³ + 6³ = 3⁵

a³ + b³ = c⁶                                   1 + β3 = a³ γ6                                 not possible (FLT)

a³ + b³ = c⁷                                    α3 + β3 = a⁴ γ7                                          4³ + 4³ = 2⁷

a³ + b² = c³                                     1 + β2/a = γ3                                          7³ + 147² = 28³

a³ + b⁴ = c³                                     1+a β4 = γ3                                            7³ + 7⁴ = 14³

a³ + b⁵ = c³                                     1 +a² β5 = γ3                                    91³ + 13⁵ = 104³

a4 + b4 = c2                                                1 + β4    = γ2/a2                 not possible (all root numbers squares)

a⁴ + b⁴ = c³                                     1 + β4    = γ2/a                         4⁴ + 4⁴ = 8³

a⁴ + b⁴ = c⁵                                     1 + β4    = aγ5                                    17⁴ + 34⁴ = 17⁵

a4 + b4 = c6                                                 1 + β4    = a2γ6                  not possible (all root numbers squares)

a³ + b⁴ = c⁵                                     1 + a β4    = a²γ6                              256³ + 64⁴ = 32⁵

 

                                                              -7-     

It is noted that the two multiplicative factors β and γ are related to ‘a’ in all cases of existing power relations and the integral values of β and γ determine the integral values of ‘a’ which stands as a common factor between (a,b,c). The power relations exist as Beal’s relation[10] only when at least one of the exponents is different .When all the exponents are same ,β and γ are unrelated to ‘a’ and become either complex or irrational   and consequently for all possible values of β and γ the three member power relation fails to exist with all integers . When β , γ and a are related with powers that are multiples of same number , the choice to make ‘a’ to be an integer becomes impossible. All the root numbers cannot be squares or cubes or like powers in three member like power relations is an alternative statement of FLT.

 

A general expression for a sum of two cubes equal to n ^th power of a number is given by [a(a³ + b³)^(n-1)/3]³ + [b(a³ + b³)^(n-1)/3]³ = (a³ + b³)ⁿ . When n

= 3. All the three root numbers cannot be whole integers..When n = 1,4,7….. (1+3x). all the root numbers can be integer but the sum cannot be a cube. It proves FLT.

 

2.  Proof-3: From Ramanujan’s Theorem

A sum of two cubes cannot be a cubes with all whole integers, but can be shown as a sum of two or more number of cubes, a³ + b³ = c³ + d³. Ramanujan gave a theorem for the sum of equal power relation. According to him, if p,q,r are three factors depending on two independent variables a,b such that p = (a+b)² – 3a², q = (a+b)² – 3ab , r = (a+b)² – 3b². and are related with two independent variables m,n by m(mq+nr)³ + n (mp + nq)³ = m(np + mq)³ + n(nq+mr)³. A typical solution is given as. (3a² +5ab – 5b²)³ + (4a² – 4ab + 6b²)³ + (5a² – 5ab - 3b²)³ = (6a² -4ab + 4b²)³ .If one of the terms in the LHS is made to be zero, the independency between a and b is vanished and consequently all other remaining terms become either complex or irrational. It proves that it is impossible to keep all the terms to be integers in any sum of two cubes equals to a cube. It is substantiated with the relations proposed by Young [9] and Vieta

[9] for a sum of three cubes equals to a cube by showing that, it cannot be converted into a relation for a sum of two cubes equals to a cube by any means.

 

 

                                                          -8-

      In any general form for a sum of two cubes equals to a sum of two other cubes one cannot convert it into a relation for a sum of two cubes equal to a cube by making one of the terms in RHS or in LHS equal to zero. This can be proved by formulating a cubical relation. For example the relation (x+ma)³ – (x- ma)³ = [(x-na)/b] + c}]³ –[(x-na)/b] –c)]³ , with b > 1 and m≠ n gives a conditional relation m³a³ + 3max² = c³ + (3c/b²)(x-na)². If c = mab² , x = (a/6n)[ m²b⁶ + (3n² – m²)].With the three independent variables, m,n and b one can construct a cubical relation b[m²b⁶ + (3n² – m²) +6mn]³ +[m²b⁶ + (3n² – m²) -6n²

– 6mnb³]³ = b[ m²b⁶ + (3n² – m²)- 6mn]³ + [m2b⁶ + (3n² – m²) - 6n² + 6mnb³]³. If one of the terms in RHS or in LHS is equal to zero, then all other terms turn into either complex or irrational.

This is true for any exponents ≥ 3. For example, Gerardin [9] proposed a relation n = 4 (a+3a² -2a³ +a⁵ +a⁷]⁴ + [1 + a² -2a³ -3a⁵ + a⁶]⁴ = [a -3a² -2a³ +a⁵ +a⁷]⁴

+ [ 1 + a² -2a³ + 3a⁵ + a⁶]⁴ If any one of the terms in RHS or in LHS is zero, then all other terms become either complex or irrational which proves FLT.

The number theory of cubical relations predicts the same result . 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 givesa³   + b³ = (6m+n)³ - 3ab (6m+n) =3c²n + 3cn²

+ n^3. Solving for c, c = - (n/2) + (1/6n) √ [ 288 m³n + n⁴ + 144 m² n² + 24 m n² – 4abn(6m+n) }. By giving suitable integral values for m,n and (a+b) one can instantaneously construct a³ + b³ + c³ = d³. This method is well suitable to get a³

+ b³ + c³ = (c+n)³ . It predicts that for every value of a (= m), there is a numeral relation a³ + b³ + c³ = (c+1)³ . 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

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

m=1                                       m=1

 

 

 

 

                                                         not exist when either a or b becomes zero. When n= 0 or 1, the relation simply reduces to 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.

 

FLT can be proved by showing that the n^th power of a number can be expressed as a sum of two powers where at least one of the exponents is different from n. Let a³ = cⁿ – bⁿ . All such three members like or multi-power relation

3.   Conclusion

Fermat Last Theorem is proved with known arithmetic axioms by three different ways- by showing the non –existence of any real solutions using the prime factors method , impossibility of converting sum of two like powers of same or different numbers into a power of a number with same exponent which allows multi-power relations and the impossibility of four members into three members relation .They are supposed to be the simplest of all proofs published.

 

References:

1.Simon Singh, Fermat’s  Enigma ,Fourth estate Ltd, (1997) 

2.Takeshi Saito, Fermat’s Laat Theorem: The proof, AMS, (2014)

3.Edward,H.M.Fermat’s Last Theorem – a genetic introduction to the Number Theory, Springer,(1997)

4.S.Wagstaff, Fermat's Last Theorem is true for any exponent less than 1000000 (Abstract), AMS Notices, No. 167  p. A-53, 244, (1976)  

5.Andrew Wills , Modular elliptic curves and Fermat’s Last theorem , Annals of Math, 141, 443-551, (1995)

6.Gerd Falting, The Proof of Fermat’s last Theorem by R.Taylor and A.Wills ,AMS,  42(7), 743-746. (1995)

7. M.Meyyappan ,A versatile proof of Fermat’s Last theorem,  http://www.ijmttjournal.org / 45(1) 503, (2017) 

8.M.Meyyappan, Resolving Fermat’s Last Theorem by Prime factor Method and proof in 5 steps http://www.ijmttjournal.org/ 46(1),  50, (2017).   

9. Tito Piezas ,A collection of Algebraic identities (tpiezas@gmail.com), (2010)

10.. L.Torres di Gregorio, Proof for the Beal conjecture and a new proof for Fermat’s last theorem, Pure & appl.Math.J, 2(5),149-155. (2013)

 

 

.