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The structure of Fermat's triples can be immensely useful in finding a simple proof of
Fermata's Last Theorem. In this contribution, the structure of Fermat's triples 1s
determined using Fermat's little theorem producing a new lower bound for the triples.
Fermat's last theorem can be stated as the equation
zn =yn +xn (x,y)=1 (1)
has no non-trivial integral solutions for x,y,z any prime n > 2 .Due to the famous work
of Germain Sophie , if we assume the existence of non trivial integral triples (x,y, z)
for any prime n > 2 satisfying (1),there may be two kinds of solutions , namely, one of
(x,y,z) is divisible by n and none of (x,y,z) is divisible by n ,and the well known
lower bound for positive x, y, z is n , that is, if x is the least, then x > n [ 1].
Let us first consider the triples satisfying xyz :;t: O(modn). Then (z - x) = y��, z -y = x��,
x + y = z�� , where xa, Ya, za are the factors of x, y, z respectively.
z�� -y�� - x�� = x + y - (z - x) - (z -y) = 2(x + y - z) (2)
x+y - z=z�� - za��=za(z��-1 - ��)=za(z��-1 -1+1 - ��) (3)
,where z=za��·Since x+y - z=O(modn) ,which follows from (1) and Fermat's little
theorem, and also z��-l -1 = O(modn) .Hence 1 - �� = O(modn)and �� = (nk + 1) , where k
is an integer which is non negative since �� :;t: 0 .Therefore , we conclude in a similar
manner that
z = za(nk + 1)
y = Ya(nl + 1)
x = xa(nm+ 1)
where k,l, m are positive integers and xa :?: 1 , m particular. Also,
x + y- z = x - x�� = O(modn) , from which it follows at once that x.:?: n + x�� > n , which
first obtained in a different manner by Grunert in 1891[1].In this contribution , it is
shown that x very well greater than n2.
Proof.
2( X +y - z)= z�� -y; -X��= ( z - za n k) 2 - (y -ya n tr - (X - X a nm) 2 = n 2 L ( 4)
due to zn = yn + xn and hence 2(x + y - z) = n2 L.
where L is an integer, Hence, x + y - z = O(modn2) . It is easy to check that
(5)
But. x+ y-z = x-x�� and all numbers za,Ya,xa,n eo-prime to one another, and hence
(6)
and note also that z a , y a > 1 which guarantees that x is very well greater than n 2 • We
deduce that
zn-z = O(modn2)
yn-y = O(modn2)
xn -X = O(modn2)
(a)
(b)
(c)
since x + y- z = (z-nkza)-z = zn-z + n2 H, where H is an integer, from which (a)
follows, and (b) and (c) follows in a similar manner. Now it is easy to deduce that
(x + y)n-zn = O(modn3) (7)
In case of (7), one has to use the simple result that if ab *- O(mod n) and
a -b = O(mod n11) ,then an -bn = O(mod n11+1) , where n 2:: 3 is a prime.
Now assume that xyz = O(modn) , and suppose that y = O(modn) , for example.
Then, since y is of the nnflanyn, it follows from the above result thatz-x = nnfl-Ian,
where a may takes positive values including a= 1. Now the equation (4)takes the form
z�� -nnfl-Ian -x�� = 2(x+ y-z) (8)
Now, since x + y- z = O(mod n2) , it follows that z�� -x�� = O(mod n2) , and it is easy to
deduce
(9)
Hence x-5n = O(mod za.nnfla. xa) and from which we deduce that x > za .nnfla. xa ,
where fJ;;::: 2. The equations (a), (b),(c) can be obtained exactly in the same manner as
before. It is easy to understand that above equations hold even if one assumes
z = O(modn) . |
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