Hello everybody,
This is my first post in this forum. I would like to write a demonstration of the Fermat Theorem I developed in late 1993, put in a floppy and then forgot for almost 17 years. Now I found it again and want to spread all over the web to see if it is valid. If this is the case I invite you to send me a message to
g_pet@hotmail.com.
Well let's go (forgive my italianish english).
Hp: x,y belong to rationals (x,y <> 1 and 0)
n belongs to naturals; n>=3
Th: x^n + y^n <> 1
Demonstration by absurd. Let's suppose that it exists at least one couple
x and y so that x^n + y^n = 1.
x^n + y^n = 1 <=> x^n = 1 - y^n <=> x = (1- y^n)^(1/n)
since y belongs to rationals so it can be written as
y = p/q with p and q belonging to integers.
let's suppose that they are also primes between them otherwise we simplify the common factors
. From this it comes that:
x = (1 - (p^n)/(q^n))^(1/n) <=> x = (((q^n)-(p^n))^(1/n))/(abs(q))
At this point we can note that also ((q^n)-(p^n))^(1/n) will belong…
to rationals because it equals x*(abs(q)) which is itself a rational
Therefore ((q^n)-(p^n))^(1/n) has to be of the form a/b with a and b
primes between them.
((q^n)-(p^n))^(1/n) = a/b <=> q^n - p^n = (a^n)/(b^n)
From which
(q*b)^n - (p*b)^n = a^n <=> a^n + (p*b)^n = (q*b)^n
Here is the absurd. In fact the number on the right contains in its
decomposition in primes factors the factor q: now for the Euclide's
theorem on the uniqueness of the decomposition in primes factors
also the number on the left has to contain the factor q
which is
not possible because at most two cases should present, both absurd
1) p contains in his decomposition the factor q: this is against the
hypothesis that p and q are primes so this possibility has to be
thrown away
2) b contains in his decomposition the factor q: in this case also the factor
a has to contain, in his decomposition the factor q and thus a and b could
not be primes between them: also this has to be thrown away because
of the previous hypothesis
At the end the negation of the thesis is absurd and therefore it is demonstrated
Ciao
Gianluca Pettinello