Fermat asserted in the early 17 th century that it is impossible to find two cubes or two higher powers that sum to another cube or a higher power respectively. For n > 2, there are no integer solutions to the equation xn + yn = zn. Fermat claimed to have a valid proof of his conjecture which appears as one of the marginal notes in his manuscripts. As the margin was too small to accommodate the proof, he left the space without any description.Fermat made a number of similar marginal notes, all of which had been resolved satisfactorily except this one- hence the title of Fermat Last Theorem.
Euler in 1753 proved FLT for n =4. Gauss corrected his attempt for n = 3 in 1825.Legendre,in 1839 proved FLT for n =5 Lame proved it for n =7 in 18 th century.Germain added additional values of n to the proven list.D .Lehmer and E.Lehmer expanded their work in 20 th century.Joe Butler and Richard Crandall brought Computer technology to FLT and prove it for all exponents n < 4,000,000. A proof of FLT was given by R .Taylor and A.Wiles on the basis of elliptic curve and modular theory.
Nature loves symmetry but also to be always simple. In fact mathematical descriptions made by human beings to understand the nature are not exactly what the nature is. There must be a simple and natural proof for FLT with no complexity in mathematical descrption.
1.Conservation of odd-eveness in equal even power relations
The numeral relation in the form an + bn = cn is found to exist only when a,b,c and n are all whole integers provided n =1 or 2.
When n = 1 ; a+ b = c, then an + bn < cn for n > 1
when n = 2 ; a2 + b2 = c2 then a + b > c and an + bn < cn when n > 2
when n = 3; a3 + b3 = c3 then an + bn > cn when n < 3 and an + bn < cn when n > 3
In general an + bn = cn then an + bn > cn when n = 1 to n-1 and an + bn < cn when n = n+1 and all higher powers.
When n - 2; a2 + b2 = c2 and are called Pythagorus relations, where the sum of two squares is equal to a square. The set of values (a,b,c) is called Pythagorean triple.
The Pythagorean triples cannot have all possible values as they have to satisfy the Pythagorus relation.
a b c
even even even
odd odd odd
odd even even
even odd odd
odd odd even
Out of various individual possibilities, the first one will be redicible where 2 will be a common factor for all the triple.Ultimately it will reduce to the permissible form.
The second one is forbidden as the sum of two odd numbers cannot be an odd number.This is true in the case of third one .
The conservation of odd-evenness necessitates that only fourth and fifth cases are allowed.
Consider the fifth case (odd, odd, even)
(2na + 1)2 + (2nb + 1)2 = (2nc)2
where na, nb and nc may be odd or even. On expansion we have,
4(na2 + nb2 ) + 4(na + nb) + 2 = 4 nc2
2 (na2 + nb2 ) + 2(na + nb) + 1 = 2 nc2
It is found that LHS is odd while RHS is even, what ever may be the values of nx ny and nz.The non-conservation of odd-eveness demands that this type of relation is forbidden. Hence the only possibility of (a,b,c) is (odd ,even, odd).
(2na + 1)2 + (2nb)2 = (2nc + 1)2
4(na2 + nb2 ) + 4na + 1 = 4 nc2 + 4nc + 1
(na2 + nb2 ) + na = nc2 + nc
nb2 = (nc - na) (nc + na + 1)
If nb is odd ,both the factos must be odd. If nc - na is odd nc + na will also be odd or nc + na + 1 will be even. .Hence nb must necessarily be even.
We can prove that in all the equal pwer relations what ever may be the value of exponent n ,the former form is forbidden while the later one is permitted.. It is very easy to establish this fact when the exponent is even.
When n = 4
(2nx + 1)4 + (2ny + 1)4 = (2nz)4
16(nx4 + ny4 ) + 32 (nx3 + ny3) + 24(nx2 + ny2 ) + 8(nx+ ny) + 2 = 16 nc4
It is reduced to
8(nx4 + ny4 ) + 16 (nx3 + ny3) + 12(nx2 + ny2 ) + 4(nx+ ny) + 1 = 8 nc4
It is impossible due to non-conservation of odd-evenness.It shows that the sum of two odd numbers raised to any even power is always singly even,when divided by 2 gives an odd number. It implies that the only possibility of formation of triplet with even exponent is (odd,even,odd).