A versatile proof of Fermat’s last
theorem
(Dr.M.Meyyappan, Professor
of Physics (Retd), Alagappa Government Arts College, Karaikudi- 630 001)
Abstract:-
A simple and
multi-purpose proof is proposed with well known arithmetic axioms to explain
not only the Fermat’s assertion but also the non-existence of square triples
(a,b,c) satisfying an + bn = cn for all exponents including n
= 2, reason for certain allowed and forbidden relations among equal sum of like
powers and Beal conjecture (Mauldin/Tijdeman-Zagier conjecture) altogether
successfully in a single frame.
Key
words:-
Fermat’s
Last Theorem - arithmetic axioms equal power relations – Multi-power relation -
Beal conjecture.
Introduction:-
For all exponents
greater than 2 , the numerical relation in the form an + bn
= cn cannot exist when a,b and c are restricted to
have whole integers which is called Fermat’s Last Theorem (FLT) , but is true
when irrational and complex numbers are allowed for at least one or two members. After the Fermat’s
assertion, many people 1-3 attempted to prove FLT for different
values of exponent and verified with the help of number theory. After the
introduction of computer technology people attempted to support the proof of
FLT for a wide range of exponents 3 - 4,000,000 4. A proof of FLT
was given by Taylor and Wills5,6 with the help of modern concepts of
mathematics – elliptic curve and modular theory. In general there must be two
or more ways to get the same result for a given mathematical problem. Since the
law of nature is always as simple as possible, there must be more natural and
naive proof for FLT. This paper describes the attempt undertaken to disclose
the naive and accessible proof of FLT with the help of well known arithmetic
axioms and number theory. The unequivocal representation reveals its
versatility of the methodology.
Arithmetic axioms:-
1.In an + bn
= cn , c is always greater than both a and b
but cm is less than am + bm
when m < n, and greater when m
> n. This is true even when m is a fraction
2.The conservation of odd-evenness in the three member equal
sum of like powers necessitates that the allowed and irreducible set of triples
(a,b,c) must be with (odd, even, odd),
where as the apparently allowed triples with (odd, odd, even) is forbidden due
to non-conservation of odd-evenness of the relation, as a doubly even component
cannot
be balanced with a singly even component, and it is found to
be true for all exponents. This restriction is given up in multi-power
relations.
3.If a is prime in a three member equal
power relation, then c - b will invariably be equal to 1
4.The
product of any two powers to give a single power follows some rules. If ax
x
by = cz, either the base numbers must be same or some
of its power or the exponents must be same or some of its multiple. When b is
equal to aα, where α may have any integer value, then
z = x + α y and c = a. When y is
equal to αx, then z = x and c = abα .
5.In ax = by,
if a,b and the exponents are all whole integers, then a = αy and b = αx, so that ax = by = αxy
Proof of Fermat’s Last Theorem:-
In an + bn
= cn, an is equal to a product of two factors (cn/2
-
bn/2) (cn/2 + bn/2) and an/2 > cn/2 - bn/2,, one can ingeniously
separate these two factors as (cn/2
-
bn/2) = an/2 /k where k > 1 , not necessary to be an integer and
may have all possible values (axiom 1), then
(cn/2 + bn/2) = k an/2 . Solving for bn
and cn in terms of an we get bn
= an [ (k2 – l) /2k]2
and cn = an [( k2 + 1) /
2k]2. In the same manner one can solve for an and cn
in terms of bn also, which gives mathematically identical
result. an + bn
= cn is then rearranged with its internal
constituents as [(2k)2/na]n + [(k2-1)2/n a]n = [ (k2
+ 1)2/n a]n where ‘a’ is found to be a common prime
factor of the triple with an inherent
condition 1 + [ (k2 – l) /2k]2
= [( k2 + 1) / 2k]2 .
Mathematically speaking such equal power relations have a common prime factor,
since any two members can be expressed in terms of the other member. They are
reducible due to this common prime factor. An irreducible set of triple for an
exponent n will not have any common prime factor at all. Besides that at least
two members will be either irrational or complex instead of being whole
integers. But the common prime factor
remains in all the three member multi-power relations as stated in Beal
conjecture, because it cannot be reduced as done in equal power relations. When
reduced it becomes a generator of multi-power relation which will not have the
common prime factor.
Whatever may be the way by which a member is factorized with
respect to the remaining members, the internal constituents may differ, but in
general, the structure of the basic relation will be same. If a = p, a prime, then the relation
becomes, pn + [(pn
– 1)/2]2 = [(pn + 1)/2]2, where
the difference c - b = 1 (axiom-3), where no common prime factor exists. If a = pq where p < q, then (cn/2 - bn/2)
= pn and (cn/2 + bn/2)
= qn ,it gives yet another form of the above expression (22/n
pq)n + (qn – pn)2 = (qn
+ pn)2, where the common prime factor may or
may not exist.
In fact
all the three member equal or unequal power relations have the following
general form an + an
[x2] = an [y2], , where [x2]
and [y2] may have any integer or a quantum of value or fractional
value but with same denominator satisfying the condition 1 +[x2] = [y2]. For a given value of the
exponent, one can change the base numbers only. Due to this limitation, the
formation of an + (βa)n =
(γa)n is practically
impossible as both βn and γn cannot be whole integer simultaneously. By
choosing a quantum of value for [x2], one term can be converted into
a modified power as required, but it is
not possible for the other with (1 +[ x2]).Even by adding any equal
quantum of value in both side, the condition fails to support the relation. For
example, by adding q both sides, the balance of the relation will not be
affected. (anβ) + q = mn
and an ( 1 + β) + q = nx which demands 1 = nx - m x and for which no integer solutions are
available.
Any known relation with whole integers such as Pythagorean
relation can even be considered for this purpose. It can be used as a key to
solve problems related to equal sums of like and unlike powers with same or
different number of terms in each side. The scope of this paper is restricted
only to three member equal and unequal power relations.
General Proof of FLT:-
Consider a basic
relation (2k)2 + (k2-1)2 = (k2
+ 1)2,To convert it into an equal power relation with an
exponent n, the requirements are k2 = an/4, (k2 -1)2
= bn and (k2 + 1)2
= cn , with a condition cn/2 - bn/2 =
2.This is valid only when n = 2 for whole integral values of c and b.
When an + an [x2] = an [y2] represents the equal power relation, the
condition is 1+ bn = cn . It is found that the relation becomes
distinct when n ≤ 2 and n ≥ 3. When n = 2, the terms within each bracket will
be whole integral number and will not have any common prime factor in its
irreducible form and the condition reduces to c – b = 2, which is practically
possible. In the other case the condition is possible when b and c are
fractions with equal denominator. When n is greater than 2, the condition
becomes unrealistic and all the members cannot be expressed as whole integral
number simultaneously.
It is found that the
terms inside each of the brackets can be made to be whole integer only when n =
2, which is the case of Pythagorean triples. When n ≥ 3 , not all the three
members will be whole integers simultaneously and one or more members will
invariably be irrational or complex. It explains the argumentative support put
forwarded by Pierre de Fermat for cubical and higher power relations and stands
as valid proof of FLT.
Non existence of square triples
All the members of a
triple (a,b,c) corresponding to any exponent n cannot be squares. It is true
even when the exponent n = 2.In fact it is a direct consequence of FLT. The
non-existence of square triples can also be taken as an indication of FLT. If all
the members of
a triple are squares (a2,b2,c2) for an exponent n, then (a,b,c) would be a triple for the exponent
2n.If square triples exist for the exponent 2 than a fourth power must be equal
to a sum of two fourth powers which is against the argument of the previous
section. In an + (a[x2]1/n)n = (a[y2]1/n)n,,
if a is a square number say α2 , the other two members will be squares
only if [x2] = α2n and [y2] = β2n, which gives again an
impossible condition β2n -
α2n = 1. It is not fulfilled even when n = 2.It predicts that all
the members can be squares when the exponent is ½. The general form of the
possible solution is given by 11/2 + (n2)1/2 = [(n+1)2]1/2. From
(2k)2 + (k2-1)2 = (k2
+ 1)2 we can arrive at
the same conclusion. The condition cn – bn = 1 does not permit square triples for any
exponent.
Forbidden power relations:-,
In general equal sum
of like powers is possible with equal number of terms in each side of the
expression for all exponents. Let us restrict our interest to an + bn
= cn +
dn .The basic relation to explain its characteristics is [x1]2
+[ y2] 2 = [x2]2 + [y1]2
where 1 + [x1]2 = [y1
]2 and 1 + [x2]2
= [y2]2. In
terms of independent variables k, k’, m and n it can be expressed as [ (k2
– m2) /2mk]2 + [( k’2 + n2)
/2nk’]2 = [ (k’2 – n2)
/2nk’]2 + [( k2 + m2) /2mk]2.
This can also be expressed as (a+b)2 + (c-d)2 = (a-b)2 + (c+d)2 with a condition ab = cd. For nth
power relation each term must be nth power of an integer. If the two
terms in one side of the relation are taken as an + bn , then the terms in the other
side will be (an +/- 1) and (bn -/+ 1). By
adding and subtracting a suitable number without affecting the balance, both
the term in the other side can be made to be nth power of different
numbers respectively.
But there are some
restrictions in the case of unequal number of like powers in each side of the
relation. Fermat’s assertion, a sum of two nth powers greater than
equal to 3 equal to nth power of whole numbers is an unique
restriction. This can also be proved algebraically (Annexure). The restrictions
are found to be different for different exponents. For example, a sum three squares or
three cubes can be shown to be a square or cube respectively
(4kk’)2 +
[2k’(k2 – 1)]2 +
[(k2 + 1)(k’2 +
1)]2 = [(k2+ 1)(k’2
+ 1)]2
where 1 + [(k2 -1)
/2k]2 = [(k2 + 1) /2k]2 and 1 +
[(k’2 -1) /2k’]2 = [(k’2 + 1) /2k’]2
But no such expression exists for the exponents n = 4 and
5.However a sum of four fourth or fifth powers equalling a fourth or fifth
power respectively exists. It is the
self adjustment that determines the possibility of formation of equal sum of
like powers balanced with equal and unequal number of terms.
Considering the
case where a square is equal to a sum of three squares. [y2]2
= (1 +[ x2])2
= 1 + [x2]2 + 2[x2]. To make
2[x2] to be a square, [x2] must be equal to 2 α2.
For
cubical relation [y2]3 = (1 +[ x2])3
= 1 + [x2]3 + 3[x2][y2], and 3[x2][y2]
to be a cube, it must be equal to 27α3 . The roots of the quadratic equation x2 (1+
x2) = 9 α3 is
given by x2 = [-1 ± √ (1+ 36
α3 )]/2, when α = 2, we get x2 = 8 or – 9 which gives a
numeral relation 93 = 13
+ 63 + 83.
Non-existence of equal sum of like powers in the form a4 = b4
+ c4 + d4 conjectured by Euler can be explained
successfully with the same fundamental expression- (c2)2
= ( a2 + b2)2 = a4 + b4 +
2a2b2 ,where 2a2b2 cannot be
expressed as a fourth power of a whole number with all possible values of a and
b.
Beal conjecture
Even though the
conditional equal power relation with three members is unachievable for all
exponents n ≥ 3, it is found that multi-power relations are obtainable. According to Beal conjecture7,8,9
when the base numbers a,b and c and the exponents x,y and z greater than 2 in ax + by = cz are all positive, then a,b,c
will invariably have a common prime factor. The secret behind the multi-power
relations is making up a power of an integer from the product of two terms - a
power of an integer am and a number of any kind [x2]
and [y2] respectively obeying
the condition 1 + [x2] = [y2] .This can be accomplished
under certain ways. The product am . [x2] becomes am
+ n when [x2] = an
, where the base number is unchanged but the exponent gets changed. The
product is equal to (aα)m when [x2] = αm where
the exponent remains same but the base number is changed. It is equal to (aα)m
+ n when [x2] = an
αn + m where both the
base number and exponent are changed. It is (aα)n when [x2]
= an - m αn where n > m
and it is equal to αn / (am – n) where m > n . In all the above cases, the base number of the product is some
multiples of the base number in the multiplicand, i.e., the base of the
multiplicand preserves the base of the product as its generic character. The
product can be made to be equal to βn which may miss the generic
character when [x2] = βn/am. In
such cases both the base and exponent get changed. The acceptable multi-power
relations are completely determined by the simultaneous fulfilment of the
requirement by both the products am [x2] and am [y2]
satisfying the condition 1 + [x2]
= [y2].
Even though am + bm = cm is not possible when a,b,c are all whole
integers and m ≥ 3 , multi-power relation in the form ax + by = cz is possible for certain permissible
values of x,y,z. Like equal power
relations, all the three member multi-power relations have the same basic
structure ax + ax[(k2 -1)/2k]2 = ax[(k2
+1)/2k]2, where a and k are two independent variables. This can be converted into a multi-power
relation by properly choosing the values of [x2] and [y2]
and converting the products to a power of an integer. For example ax + ax + m = ax
+ n is not possible
with same base number since 1 + am ≠ an for all possible values of a,m and n. But ax
+ ax + m = (aα)x is possible due to the soluble
condition 1 + am = αx. When a = 2,α = 3, m = 3 and x = 2,
we get 22 + 25 = 62. Similarly ax +
(aα)x = ax + m is allowed due to
attainable condition 1 + αx = am which gives a multi-power relation 33 +
63 = 35.
From the general form of the three
member equal power relation, one can derive one or more conditions for any
particular requirement. Its numeral form can be used as a generator for the
development of such multi-power relations. In fact three member multi-power
relation is not possible for all possible set of exponents due to insoluble
condition. Table.1 shows some of the allowed and forbidden 3 member multi-power
relations with a typical numerical example.
Table.1 Allowed and forbidden three member multi-power
relations
ax
+ by = cz conditions numerical example
___________________________________________________________________________
an
+ (aα)n = (aβ)n 1 + αn =
βn not
possible (FLT)
___________________________________________________________________________
a3 + (aα)3
= (aβ)4 1 + α3 = aβ4 93 + 183 =
94
(aα)3 + (aβ)3 = a4 α3 + β3 = a 703+
1053 = 354
a3 + (aα)4 = (aβ)3 1 + a α4 = β3 73 +
74 = 143
(aα)3 + a4
= (aβ)3 α3
+ a = β3
383 + 194 =
573
a3 + (aα)3
= (aβ)5 1 + α3
= a2β5 33 + 63 = 35
(aα)3 + (aβ)3 = a5 α3 +
β3 = a2
83 + 83 =
45
a5 + (aα)3
= (aβ)3 a2 + α3 = β3
135 + 913 = 1043
a5 + α3
= β3 a5
= (β – α)(β2+ βα +α2) not possible
a3 + (aα)5 = (aβ)3 1 + a2
α5 = β3 913
+ 135 = 1043
a3 + (aα)4 = (aβ)5 1 = a(a β5 – α4) not possible
In fact multi-power relations can be obtained directly from
the generator- any simple numeral expression, by multiplying each term with a
common multiplier, which provides a common prime factor to the relation. For
example, the condition 1 + 1 = 2 when multiplied by 2m gives 2m
+ 2m = 2m + 1 ,and
the common multiplier 22m gives
4m + 4m = 22m + 1 where as the condition 1 + 7 = 23 gives 73 + 74 =
143 with a common multiplier
73. an + bn + 1 = cn is the
-7-
most common type of multi-power relation. The general form of
this type of relation is given by (2n -1)n + (2n -1)n
+ 1 = [2(2n – 1)]n
A generator can provide many multi-power
relations each with different multipliers. It is exemplified with few
examples
1 + by = cz
→ 1 + 23 = 32
Only certain common multiplier alone can make each term into
a power of an integer. 1 is very convenient member as it is the only
multiplicand to make product with multiplier of any form.
In the above case
either bz or cy or
αzy will be the convenient
multipliers.
[1 + by = cz]
x bz = bz + by + z = (bc)z → 22
+ 25 = 62
[1 + by = cz]
x cy = cy + (cb)y
= cy + z → 33 + 63 = 35
[1 + by = cz]
x αzy = αzy + (αzb)y = (αyc)z
→ 76 + (98)3 = (1029)2
Similarly, ax + bx
= c → 13 + 33 = 28, multiplying with the common multiplier cx
[ax + bx = c] x cx = (ac)x + (bc)x = cx+1 → 283 + 743 = 284
with common multiplier 283
→7843 + 23523 = 287
with common multiplier 286
If ax + by
= cz itself is taken as generator, then one
of the common multipliers will be αxyz
(ax + by
= cz ) x αxyz
= (aαyz)x + (bαxz)y
= (cαxy)z
The generators usually do not have common prime factor, but
will have in its reducible form. Hence the generators may or may not have a
common prime factor depending upon its reducibility. In fact the generators are
irreducible form of multi-power relations. When multiplied with a common
multiplier to get a multi-power relation, it acquires a common prime factor.
One can show that the common prime factor is compulsory for
all the three member multi-power relations but vanishes in all the three member
equal power relations .The common prime factor is not a compulsory requirement
in multi-power relations with at least one member is a square or more than
three members. The non-existence of common factor in equal power relations for
all exponents n ≥ 3 can also be taken as an indirect proof of FLT.
Exempted multi-power relations
Beal assertion is not a necessary condition in all
multi-power relations with more than three members or one of the members is a
square. e.g., ( with more than three members 94 = 84 + 74
+ 43 ; 35 = 43 + 23
+ 23 + 13 ; 54
= 34 + 25 + 83) and ( with one member is
a square 83 = 132 + 73; 1053 = 1812
+ 1043 ; 174 = 1612 + 2402).
When one of the members is a square, then the three member multi-power relation
becomes
a2 + a2
[(k2 – 1)/2k]2 = a2 [(k2 + 1)/2k]2
a2 + ay
[1/ay – 2] [(k2
– 1)/2k]2 = az
[1/az – 2] [(k2 + 1)/2k]2
[1/ay – 2] [(k2 – 1)/2k]2 =
αy and [1/az – 2]
[(k2 + 1)/2k]2 = βz from which the required condition can be
derived as
1 = (az – 2)
βz - (ay – 2) αy
= ay – 2[ az – y βz – αy ]
which has no integral solution.
When it is converted into a form a2 + b3 = c3 , the common factor is either
lost or changed by the denominator of [x2] and [y2] . When the common multiplier is some
multiples of the common denominator of the fraction representing [x2]
and [y2] , the multi-power relation may lose its common prime
factor. That is what is found in multi-power relation with one member is a
square.
Like three member equal power
relations, in multi-power relations also all the members cannot be squares or
any equal higher powers .
a2x + a2x [x2] = a2x [y2] .Assuming x > y > z, one can construct
a multi-power relation as
a2x + a2y(a2x -2y [x2])
= a2z (a2x -2 z [y2]) → ax
+ (aβ)2y = (aγ)2z
with a condition 1 + β2y/(a2x –2y) = γ2z
/(a2x –2z).Since the denominators are different βy
must be divisible by (ax –y) and γz by (ax –z) which gives a modified
condition as 1 + m2 = n2 , which has no practical
solution for any integer values of m and n. Even by self adjustment also, one
cannot make it to be true.
Acknowledgement
The author is
grateful to Pierre de Fermat who gave an open problem to think over at least if
not to try to solve by people of all age group. This proof is dedicated to
Mahatma Gandhi, a man I love most.
Reference:-
1.Simon Singh, ‘Fermat’s Last Theorem’ (1997) Fourth Estate
Ltd.
2.Jackson,A. ‘Fermat’s Enigma ‘ (1997) Review, AMS.
3.Edward,H.M.,”Fermat’s Last Theorem – a genetic introduction
to the Number Theory’ (1997),Springer.
4.Wagstaff,S., AMS Notices No.167,(1976)p A-53,244
5.Andrew Wills, ‘Modular elliptic curve and Fermat’s last
theorem’ (1995) Annals of Maths., 141, 443-551.
6.Gerd Faltings., “
The proof of Fermat’s Last Theorem by R.Taylor and A.Wills” AMS,1995,42
(7),743-746.
7.Daniel Mauldin,R., ‘A generalization of Fermat’s Last
theorem: The Beal conjecture and prize problem’ (1997) , American Mathematical
Society, 44,1436-39.
8, Golden G.Nyambuya, ‘A simple and general proof of Beal
conjecture’ (2014) Adv.in Pure. Maths., 4,518-521.
9.wimhessellink.nl/fermat/WilesEnglish.htm
________________________________________________________
Annexure.
The algebraic method
also confirms FLT. In the case of cubical expression,
a3 + b3 = (a+b)(a2 - ab + b2) =
c3
Since a+b > c, a+b= kc where k > 1 and a2 - ab + b2 = c2 /k, Solving for a and b
a = (kc/2) – (ck/6)[
(12/k3) – 3]1/2 and
b = (kc/2) + (ck/6)[ (12/k3)
– 3]1/2
when k =1
, the second term becomes c/2 and a = 0, b = c. For all values k > 1, the
second term becomes irrational or complex.
__________________________________________________________________________________xv
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