Abstract:
The primitive Pythagorean triples (x, y, z) are now well understood [1]. However, we
believe that a closer look at the solution is needed along new directions to understand
the terrible difficulty in giving a simple proof for the Fermat’s last theorem. Keeping this
fact in mind we look at the solutions of z 2 = y 2 + x2 , (x, y) = 1 in the following manner.
(x, y, z) is a primitive Pythagorean triple if and only if
x2 + y 2 = z 2 , (x, y) = 1
(1)It is obvious that one of (x, y, z) is even and it can be shown that z is never even by
using (1) and substituting z = y + p, p ³ 1, in it.
Now either x or y is even. If we suppose that y is even, z 2 - x2 = y 2 and then it follows
that z - x = 22b -1 or z - x = 22b -1a 2 where a ,b ³ 1 and are integers. The following
are examples for the justification of our point.
2 2 2 3
2 2 2
13 12 5 , 2 , 2
17 15 8 , 1 , 2
= + = - =
= + = - =
z x
z x
b
b
1132 = 1122 +152 ,b = 1,a = 7 z - x = 2´ 72
Now we apply the mean value theorem of the form
a2 - b2 = 2(a - b)x where a <x < b , to the expression z 2 - x2 , to obtain
z 2 - x2 = 2.22b -1 a 2x since z 2 - x2 = (z - x)(z + x)
It follows that ( )( )
2
2 2 2. z x
z x z x
- = - +
It is clear that 2.22b -1a 2 or 2(z - x) is a perfect square and since ( )( )
2
2 2. z x
y z x
= - + it
follows that =x +
2
z x
is a perfect square.
Therefore, in case of any primitive triple (x, y, z) of z 2 = y 2 + x2 , the mean value
theorem is manifested in the form
a2 - b2 = 2(a - b)x where x is a perfect square b <x < a .
Now we point out the following conjecture. Suppose that z, x > n for any prime n ³ 3 .
Then, z n - xn = n (z - x)x n-1 by the mean value theorem and we conjecture that x is
irrational when z - x =a nnbn-1 .
