Posted by: jimmypa | November 16, 2012

From Fermat’s little theorem to Fermat’s Last Theorem

From Fermat’s little theorem to Fermat’s Last Theorem

Fermat’s little theorem states that if p is a prime number, then for
any integer a, the number ap-a is an integer multiple of p.

ap≡a(mod p)

Assume that a , b , c naturals and p prime and 0<a≤b<c and a<p

ap≡a(mod p)

bp≡b(mod p)

cp≡c(mod p)

Let’s assume that ap+bp=cp

We can say

a + px + b + py = c + pz with x,y,z positive integers

a + b – c = p(z – x – y)

If a + b – c ≤ 0 we have a + b ≤ c

we have (a+b)p≤cp and using binomial theorem

ap+bp<(a+b)p=ap+…+bp≤cp

In conclusion the original assumption that ap+bp=cp is false .

If a + b – c>0 because 0<a≤b<c implies b – c < 0 ,0 < a + b – c < a < p

then a + b – c ≥ 1 and because a + b – c < a implies a ≥ 2 and because a < p implies p > 2. *
If a+b -c<p and if a + b – c = p(z – x – y)
implies 0≤z – x – y<1 ,
because x,y,z are positive integers z – x – y=0
which implies
(1) a + b – c = 0 and a + b = c , raising to power p ,
ap+bp<(a+b)p=cp
Again the assumption that ap+bp=cpis false .
(if z – x – y < 0 , a + b – c < 0 and we are in previous case
and if z – x – y ≥1 , a + b – c <a < p < p(z – x – y)
and then a + b – c ≠ p(z – x – y) and ap+bp=cp is false )
We can’t find naturals 0<a≤b<c with a<p with p odd
prime to satisfy ap+bp=cp.

We can extend this to a , b , c rational numbers ,
0<a≤b<c and a<p .
If ap≠ cp– bp is true for a , b , c ,naturals a < p , is true for a rational , a < p and b , c naturals because cp– bp is natural.
We can divide ap≠ cp– bp with kp , k rational k > 1 and note (a/k) = q ,
qp≠ (c/k)p– (b/k)p with q rational q < p.

Let’s pick d positive integer , p ≤ d , d≤b < c and
assume that dp+bp=cp .
We can find k rational number such d/k < p and we have
(d/k)p + (b/k)p = (c/k)p which is
false of course because d/k < p .

If p is not prime we can write p as a product of prime factors .
Assume that q composite , q = rp with p prime we can write
aq = arp = (ar)p,
which implies that we can’t find positive integers to satisfy
ap+bp=cp with p > 2.

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Responses

  1. You need not try to prove it anymore. There was an elementary solution at

    http://mathforum.org/kb/servlet/JiveServlet/download/13-2413846-7919625-781687/Did%20Fermat%20prove%20FLT.pdf

  2. Oh my goodness! Impressive article dude! Thanks, However I am
    having problems with your RSS. I don’t know why I can’t join it.
    Is there anybody getting identical RSS problems?
    Anyone that knows the solution can you kindly respond? Thanx!!


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