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Fermat's last theorem can be stated as that the equation
zn =yn +xn,(x,y)=1 (A)
has no non-trivial integral solutions for (x, y, z) except for n=2, and therefore we have
carefully examined all primitive Pythagorean triples [1],[2] and we solved Pythagoras'
equation analytically resulting in a new generators for Primitive Pythagorean triples
and a simple conjecture which is explained and proved for n = 3 in the following.
1. Conjecture
When y is divisible by 2, all primitive Pythagorean triples (x,y,z) are related by
z 2 -x2 = y 2=2.(z-x)(x+Bh) (1)
where h = z - x = 2 21H a 2 and e = .I_ . This resembles the Mean value theorem
2
f(z)- f(x) = (z- x).f'(r;), where f(x) = x 2 , / (r;) = 2.r; and r; = z + x ,which is a
2
perfect square. If (A) has a non trivial integral solutions for a prime n-:;; 2, then there
may be such solutions that one of x, y, z is divisible by n which is well known .Then the
equation (A) can be put into the form
(2)
with z- x = nf3n-Jan assuming that y is divisible by n ,where all letters except r; stand
for integers .. This also resembles the Mean vale theorem
f(z)- f(x) = (z- x)./ (r;) withf(x) = xn. It is conjectured [1]that if (A) is true, then
rn = r;n-l does not hold witli integral y, r; except for n = 2 .
2.Proof of the Conjecture
In this contribution, the conjecture in the previous section is proved for n = 3. Proof is
based on the following lemmas.
2.1 Lemma
If
has an integral solution for x,y,z ,then one ofx, y, z is divisible by 3.
The proof of this lemma is simple and it is assumed without proof.
(3)
Since the above equation holds for - x,-y,-z, without loss of generality, one may
assume that y is divisible by 3.
2.2 Lemma
If y is divisible by 3, then z- x = 3'11 1 a~ ,where a is an integer including ± I. The
proof of this lemma is exactly the same as as in the case of analytic solution of the
Pythagoras" equation for primitive Pythagorean triples and is also assumed without
proof.
~ow, (3) takes the form
(4)
2.3 Lemma
If (cd) = l=(hJ). then it follmvs from the Fermat's little theorem that a' ±h1 IS
divisible by 3 and since a' ±h' = (a±h)((a±h)2 ±3ah),the least power of 3 that
divides a~ ± h' is 2.
Substituting z-x=3'13 1a' m (4),oneobtains
:> - 31>{1-1 6 + '11/1-1 + 2 )I - a .) .X X (5)
and (5) can readily he put into the form
4 r' =36fl3a6 +(2x+ 3 ,;r-la')2 = 4(
If ¢ is an integer it can be expressed as x +!',where ,u IS an integer,
follovvs from (6) that
(6)
and then it
3 1' fH a 6 = ( 4 x + 2 p + 3 ' 11 1 a 1 )( 2 JL - 3 ' 11 -1 a 3 ) (7)
Let us assume a is a prime for simplicity .If (2p- 3111 - 1 a') is not equal to I , then it
cannot be 3611 1 a 6 since then 3x +x-I+ 2p + 33.8-l a 1 = 0 from which and (5) it
follows that(3.x) :t:: I Therefore 2p- 3311 -1 a 3 is equal to a 6 or 31'!1-1 and hence
or
or
2p-3'11-1 a' = l
and (c) gi \"CS
3611 1a' =4x+2.3'11 1a 1 +1
(a)
(b)
(c)
(d)
Since x-I or x + 1 is divisible by 32 due to (5) and since a3 + h3 is divisible by
32 we conclude that (a) or (b) or (d) is never satisfied since (3,x) :f:: 1. Hence our
assumption that s and y are integers never holds . If a is a composite number it can
be expressed as a product of primes and the proof of the conjecture follows in the similar
manner as above. |
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