The Last Theorem of Pierre de Fermat, in elementary way

In this paper the author works only through the factorizzation in factors, with the proceeding for absurd, that is if x ,y, z are prime among them, under the hypothesis that the tern of integers (x, y, z) were a solution of the equation x n + y n = z n Then he obtains that the first and the second term of an equivalent relation are odd (the first) and even (the second). Three cases are separated: 1) n is power of 2; 2) n is odd; 3) n is product of a power of 2 for an odd number. For a better understanding also the Pythagorean set of three numbers is reported.


Introduction
Pierre de Fermat,(born at Beaumont de Lomagne in 1601 and dead at Castres in 1665), belonging to "noblesse de robe ",who conferred him the noble symbol "de" in front of his surname, through a mirabilis proceedings, proved, about 1637, that the equation x n + y n = z n has not any integer positive solutions, when "n" is more than two. Unfortunately his demonstration has not passed on to posterity. In fact, in the first edition, made by Bachet de Meziriac, of the book "Arithmetica" about the works of Diophanto, a Great Greek mathematician of the third century before Christ, Pierre de Fermat, did not think to quote it because it was too long, but in the margin of the same pages he referred to have a mirabilis proceeding of proof. Nor did it the son Samuel, who, in 1670, edited the reprint of the book on Diophanto titled Varia Opera Mathematica, even if he added some notes, he did not report such a demonstration. During these four centuries the most renowned mathematicians have tried, without success, such demonstrations, for this they questioned the assertion of Fermat not to have succeeded. J. Andrew Wiles tried the Theorem but in complicated way. This work is the proof that the elementary demonstration is there and I am sure that Pierre had formulated it in a plainer way. In the short demonstration, that has been reported here, the various passages have an elementary nature, therefore they were perhaps also in the mirabilis proceeding of Fermat; so, this work could be considered on archaeological type. However it is probable that Pierre has used his knowledge on the effective powers of the numbers also with many digits, or those on the spheres and iperspheres, or those on the parabolas of the type y = z n and y = ( z -h) n , as he was great at geometry.

Preliminary observation
Even if it is obvious and perhaps, for a mathematician as Pierre de Fermat really trivial and not important for the proof, here one introduces the Zero Theorem: For the possible integer positive solutions of the equation x n + y n -z n (1) must be Proof: Just to give everything more understanding also to people, who have no familiarity with numbers, one reports the demonstration. We suppose x, y, z, prime among them, because, if (x, y, z) is a solution, also (hx, hy, hz) is still a solution and on the contrary: we observe that if two of three values are not prime, that is they have a common factor, then also the third has the same factor. Then x, y, z cannot be all three odd at the same time: one of the three numbers has to be even. Therefore x + y -z is even. Immediately we notice that only for n=1 appear x + y -z = 0 For the predicate you have only to prove that, if one supposes x + y -z = 2 , then the equation

(α)
x n  y n  z n n  3 has no integer positive solution (x, y, z). The proof is symmetric with respect to x and y. Let If it is x=1,therefore y=z+1 and one should not have written because it is impossible. If it is x=2, therefore y=z and one should not have written 2 n  z n  z n because it is impossible. If it is x=3, therefore y=z-1 and the equation 3 n  ( z  1) n  z n z  5 and n  3 is impossible , because we have obviously For example, when z=5, it is 3 n  4 n  5 n n  3 .
Clearly (z-1)ⁿ increases less than zⁿ, whether z increases or n increases: one concludes that (β) is Therefore it is allowed the condition   4 n  ( z  2) n  z n z  5 and n  3 For example, when z=5, we have 4 n  3 n  5 n n  3 .
Clearly (z-2)ⁿ increases less than zⁿ, whether z increases or n increases: one concludes that (δ) is For h=3 the equation 5 n  ( z  3) n  z n z  6 and n  3 has got no solution since (ε) 5 n  ( z  3) n  z n z  6 and n  3 .
For example, when z=6, we have 5 n  3 n  6 n n  3 .
Clearly (z-3)ⁿ increases less than zⁿ, whether z increases or n increases: one concludes that (ε) is true z  6 . Generally (z-h)ⁿ increases less than zⁿ, whether z increases or n increases: one concludes that (γ) is true z  5 and n  3 , and the equation (α) has got no primitive integer positive solution.
So, the search for any possible solutions of x n  y n  z n is restricted to the case q.e.d. It is easy to proof, but it is not necessary and here we do not quote, that the search for any possible solutions of (1) is restricted to the case Generally, we do not quote the case also if it is probably the beginning of the way of Pierre de Fermat.

The Pythagorean set of three numbers and the case n power of 2
The equation has got as solutions the Pythagorean set of three numbers. We suppose x ,y, z, prime among them, that is gcd(x, y, z)=1, because, if (x, y, z) is a solution, also (hx, hy, hz) is still solution and viceversa. We observe that if two of three values are not prime among them, that is, they have a common factor, then also the third has the same factor. Then x, y, z cannot be all three odd at the same time: only one of the three numbers has to be even.  Proof: In the hypothesis that x ,y, z are prime among them, x and y cannot be both even: we suppose y odd and let we have that k divides y², and its factors are square, that is, k takes up in the form of In fact, if a factor of k is not square and divides y, then it divides also 2x, that is , x or 2, against the hypothesis that x, y, z are prime among them and y odd.
Therefore we have that also 2x+u² is a square and also it is v 2  2 x  u 2 with u and v prime among them.
In the end q.e.d.
We note that (if) y is odd, then x is even and z is odd, because v-u and v+u are even and Second Theorem: The equation (2) has no integer positive solution.
Proof: We could suppose, that (x, y, z) is a solution with x, y, z prime among them and y odd. The equation can be written so that (x², y², z²), will be a Pythagorean set of three numbers, that is with u and v prime among them, then is even, but y 2 and z 2 are odd, that is x even, y and z odd. So, that is (x, z, v) and (x, u, z) are Pythagorean set of three numbers, as x is even, so z, u, v are odd and prime among them, then exist a, b prime among them and odd, so c, d and we have:

The Bulletin of Society for Mathematical Services and Standards Vol. 4 25
The proof is based on the first digit, or digit of the units, of a number. We observe that y 2  uv is square and since u, v are odd and prime among them, so u and v are square, that is We indicate with fd(x) the first digit of a number x and we note that the first digit of the fourth powers of an odd number is 1 or 5 , whereas of an even number is 0 or 6 and that the first digit of the second powers of an odd number is 1 or 9 or 5, whereas of an even number is 4 or 6 or 0.
We have: But u and v are prime among them, therefore fd ( In the I case, two numbers, with the first digit equal 1, have the difference multiple of 10, so we can write: x  10 s t , y  10Y 1 , z  y  m10 r  10Y 1  m10 r , with fd(t)>0, we have r=4s, because, by we have we can write again This is absurd because and fd(second term) = 4m, that is

= 4m or 5 = 4m
but it is impossible, because the second term is even and the first is 1 or 5: only if t = 0 and m = 0 the relation (5) is possible, that is but this is impossible because x is positive.
In the II case we have: and we return to the I case.
It is similar in the cases III and IV. We conclude that the equation (2) has no integer positive solution. q.e.d.

Third Theorem:
The equation x 2 n  y 2 n  z 2 n n  2 has no integer positive solution. and there are no solution for the second theorem. q.e.d.

The case 2n+1, that is the Last Theorem of Pierre de Fermat
Here is the Last Theorem of Pierre de Fermat and Andrew Wiles: The equation x 2n +1  y 2n +1  z 2n +1 (1) 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 2n+1: let y not be multiple of 2n+1.
Let z = x + k with k prime with x. It follows that Every factor of k must divide y 2n +1 , then every factor of k will be a power of exponent 2n+1, otherwise it should divide also (2n + 1) x 2 n , that is x or 2n+1, against the hypothesis; so k will be in the form of where u and v are prime among them.
This result, that is y = uv and k = u 2n +1 , can be obtained also by the relations of Waring. We observe that v is always odd because x and z cannot be even at the same time.

Volume 4
By the relation (2) , working out the term 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 2n 1 is either odd or even. At the same time, we have z 2n 1  x 2n 1  y 2n 1  ( x  y)( x 2 n  x 2 n1 y  ...  y 2 n ) . We remember that x and y are prime among them, so x+y is odd if x is even and divides z 2n+1 then the factors of z, that divide x+y , cannot divide xy, that is x 2 n  x 2 n1 y  ...  y 2 n , such factor of x 2n +1 + y 2n +1 we indicate with p 2n +1 .
One concludes: x  y  p 2n +1 z  pq , that is and z 2n +1  p 2n +1 q 2n +1 , with p and q prime among them and q odd and q 2n +1  x 2 n  x 2 n1 y  ...  y 2 n .

Finally, we have
x  pq  u 2n +1 and x  p 2n +1  uv .
We remember that when z is odd, then p and q are odd. Therefore it is

The Bulletin of Society for Mathematical Services and Standards Vol. 4 29
The factors p, q, v and u are prime among them and p+u divides z+y , so -c), but u and p are prime among them, so exists a such that c-v=ap and q-c=au or where and q  au  v  ap  u 2 n  u 2 n1 p  ...  p 2 n ; By those relations, one has: 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 t  p 2 n  q  ul and s  v  u 2 n  pl , we have: 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.

Volume 4
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.
We observe that l is the factor of x and of z-y in fact it is 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: z  y  l 2n +1 and x = lf , so f 2n +1  z 2 n  z 2 n1 y  ...  y 2 n ; with l and f prime among them and f is odd. At the same time , we have: from here, we obtain that in which p and l are prime among them; then We have: The Bulletin of Society for Mathematical Services and Standards Vol. 4 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: x  lf , y  uv , z  pq and lu  p 2 n  q , pu  f  l 2 n , pl  v  u 2 n . ( Now we consider It is: Then, we have: We observe that d is even if u is odd and p even. Finally, we have: which verify the (2), that is and p 2 n1  u 2 n1  l (l 2 n  2up), p 2 n1  l 2 n1  u(u 2 n  2lp) , u 2 n1  l 2 n1  p(p 2 n  2ul ) In addition, we conclude that since, from (8), f  l 2 n  up , v  u 2 n  lp , q  p 2 n  ul We observe that With x 1 , x 2 and x 3 odd. 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 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 with L,U 1 ,U 2 ,U 3 , P 1 , P 2 , P 3 , F 1 ,V 1 , Q 1 , C 1 , C 2 , C 3 odd The problem is symmetric with respect to a 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) ≻1 , for relations (9) that is We conclude that we can write C 3  γ    , with  1 odd.

Volume 4
By c 3  q  ul we obtain that is or this relation is an absurd because the first term,  ≻ 0 and  ≻  1 , 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.
We conclude that, in both cases, no integer positive primitive solution of (1) is possible, so no integer positive primitive solution of (1) is possible. q.e.d.

Final note:
also the equation p 2 n1  u 2 n1  l 2 n1  2upl solves the problem.
Proof: It is obvious that it is interpreted with the preceding theorems.
If we lay down and there is no solution for the proof of 2n+1, therefore (x,y,z), with is not solution of (1) ,so (X; Y; Z) of (2). q.e.d.