The Last Theorem of Fermat for n=3

: In this paper on FLT, one solves the case n=3 in elementary way, extensible to n odd. The author works only through the sole factorization in factors and with the proceeding for absurd, that is if x, y, z are prime among them, under the hypothesis that (x, y, z) are a solution, one obtains that the first and the second term of an equivalent relation are odd (the first) and even (the second).

has no integer positive solution. Proof: We could suppose that (x, y, z) is solution with x, y, z prime among them, in particular x and y cannot be both multiple of 3: let y not be multiple of 3. Let z = x + k with k prime with x. It follows that y 3 =z 3

-x 3 =(z-x)(z 2 +zx+x 2 )=k(3x 2 +3xk+k 2 )
Every factor of k must divide y 3 , then every factor of k will be a power of exponent 3, otherwise it should divide also 3x 2 , that is x or 3, against the hypothesis; so k will be in the form of k=u 3 and also v 3 =3x 2 +3xk=k 2 that is v 3 =3x 2 + 3xu 3 +u 6 (2) where u and v are prime among them. This result, that is y = uv and k=u 3 one can obtain also by the relations of Waring. We observe that v is always odd because x and z cannot be even at the same time. By the relation (2) , working out the term 3x 2 = v 3 -3xu 3 -u 6 we obtain that if x is even, then u is odd; if x is odd, then u is either even or odd. In other words one has two cases: the first one: x even, y and z odd, the second: x and y odd, z even. Therefore z = x + u 3 is either odd or even. At the same time, we have Edward Waring, professor of Cambridge University (Shrewsbury, Shropshire, 1734-Pleasley, 1798) worked out in a polynomial of x+y and xy We remember that x and y are prime among them, so x+y is odd if x even and divides z 3 then the factors of z, that divide x+y , cannot divide xy, that is One concludes: x  y= p 3 and z = pq that is z 3 p 3 q 3 with p and q prime among them and q odd and We remember that when z is odd, then p and q are odd. Therefore it is The factors p, q, v and u are prime among them and p+u divides z+y , so Bulletin of Mathematical Sciences and Applications Vol. 2 but u and p are prime among them, so exists a such that c-v=ap and q-c=au where and q -au = v + ap = u 2 -up + p 2 ; By those relations, one has: and We observe that in the relation (I) the first term is even when p is odd, in that case u l is even and also a is even. Likewise, placed We observe that l is prime with u, p, v and q; looking l: l is prime with u and p; l is prime with v: in fact, if l were not prime with v, we would have a common factor of v, but and u is prime with v; l is prime with q: so, if l were not prime with q, they would have a common factor of q, but it is and p is prime with q. We observe that all the factors of u, v, p, q, uv, up, uq, vp, vq, pq do not divide uv+pq and  u+v, p+q. For example, we proof that up does not divide uv+pq. In fact, although it is obvious, if up divides uv+pq, then that is p²q and u²v must have a common factor, against the hypothesis that pq and uv are prime among them.

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Volume 2 We observe that l is the factor of x and of z-y in fact it is

ulp  x + y -z= x -(z-y) .
Then we have l<x: in fact, if it were l=x, we would have the absurd: Now, we take: a factor of z-y, (for Waring), cannot divide zy, because z and y are prime among them; therefore, we have: with l and f prime among them and f is odd.
At the same time , we have: in which p and l are prime among them; then We have:

Bulletin of Mathematical Sciences and Applications Vol. 2 71
but, i=u because We observe that in the relation (III), the first term is even when p is odd, in that case lu is even moreover also b, as a, is even. We conclude: We observe that d is even if u is odd and p even. Finally, we have: which verify the (2), that is dividing for p 3 , we have:

Volume 2
From this verification, we are sure on the developments. For the conclusion of theorem we take again the relations: where between a , b, d at least two are even. Now, we remember: By the relations (8), we have (9) We observe that and with x 1 , x 2 and x 3 odd and we use this symbology, on the following transformations.
The problem of Fermat is symmetric with respect to x and y, therefore, for simplicity, we consider 3 two cases: the first: p odd and among l and u, l even; the second: only p even p even We go on in the first case and we use the above-mentioned symbology on the following transformations.

Let
We begin to examine the first of (8) we have two possibilities: that is or

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and We have The problem is symmetric with respect to γ and β , therefore on the cases (a), (b), we study only the case (a).
We obtain, in the case (a), Now we consider the hypothesis (a) for relations (9) this relation is an absurd because the first term, is odd and the second is even.
We conclude that, in the cases (a) and (b), no integer positive primitive solution of (1) is possible, so no integer positive primitive solution of (1) is possible. Now we consider the second case: it is similar to first one, exchanging the roles of l and p.
We deal the second case using the showed symbology on the following transformations.