prime powers?

Question

in fermats last theorem what is the neccessity to prove powers other then prime ones or the powers having only 2 as its multiple bcs we
can prove very easily that
a^(n)-b^(n)=c^(2)-d^(2) like
7^(3)-4^(3)=48^(2)-45^(2) &
x^(n)=z^(2)-y^(2) like
3^(3)=14^(2)-13^(2)
so it is impossible to write from this information
a^(n)-b^(n)=x^(n)
for n>2 & all the prime no. powers and the powers having only 2 as its multiple.

Answer 1

Suppose n is composite, say n = k * m. Then

x^n + y^n = z^n

implies

(x^k)^m + (y^k)^m = (z^k)^m

so there are numbers p = x^k, q = y^k, r = z^k satisfying

p^m + q^m = r^m.

For this reason we can do with prime numbers. The only exception is n a power of 2, in which case all prime factors are 2, and there are obviously numbers such that p^2 + q^2 = r^2. However, there is a relatively simple prove that shows that the sum of two fourth powers cannot be a sqaure (let alone a fourth power).

If p is an odd prime, the formula

x^p + y^p = z^p

is equivalent to

x^p + y^p + (-z)^p = 0.

We often generalize the problem by saying that

x^p + y^p + z^p = 0

has no non-trivial solutions for p an odd prime and x, y, z integers. Proving this is proving Fermat's Last Theorem. You can put the pluses and minuses wherever you want.

Answer 2

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Answer 3

You seem to be talking about differences of powers. Fermat is about sums of powers. There *is* a difference.

x^n + y^n = z^n has no solution triple {x,y,z} in integers if n>2

Go read Wiles proof.


Doug

Answer 4

isn't the theorem supposed to be

x^n+y^n=z^n

wher
n>2 ?

Answer 5

Partly Correct!!! But are you sure?