An elementary proof of Fermat-Wiles theorem.

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An elementary proof of Fermat-Wiles theorem.

Jamel Ghannouchi

To cite this version:

Jamel Ghannouchi. An elementary proof of Fermat-Wiles theorem.. 2014. �hal-00966814�

An elementary proof of Fermat-Wiles theorem

Jamel Ghanouchi

Ecole Supérieure des Sciences et Techniques de Tunis jamel.ghanouchi@live.com

Abstract

( MSC=11D04) We begin with Fermat equationYⁿ=Xⁿ+Zⁿand solve it.

(Keywords : Diophantine equations, Fermat equation ; Approach) Resolution of Fermat equation

Let Fermat equation :

Yⁿ=Xⁿ+Zⁿ We have

Xⁿ⁻²Y²−Yⁿ⁻²X²=AZⁿ And

Yⁿ⁻²Y²−Xⁿ⁻²X² =Yⁿ−Xⁿ=Zⁿ

IfA = 0then Xⁿ⁻⁴ = Yⁿ⁻⁴ leads, asGCD(X, Y) = 1, ton = 4. This case has been studied by Fermat, it has no solution. ThusA6= 0.

And ifA=±1then it means that both Xⁿ⁻³Y²=±^ZXⁿ +XYⁿ⁻²and

Yⁿ⁻³X²=∓^ZXⁿ +Xⁿ⁻³Y²are rationals it means thatn= 2.

We have

Xⁿ⁻²

A Y²−Yⁿ⁻²

A X²=Zⁿ=Yⁿ⁻²Y²−Xⁿ⁻²X² And we have simultaneously

(Yⁿ⁻²−Xⁿ⁻²

A )Y²= (Xⁿ⁻²−Yⁿ⁻² A )X² Or

(AYⁿ⁻²−Xⁿ⁻²)Y² = (AXⁿ⁻²−Y^p−2)X² And

(Y²+X²

A )Yⁿ⁻²= (X²+ Y² A )Xⁿ⁻² Or

(AY²+X²)Y^p−2= (AX²+Y²)Xⁿ⁻²

1

We have four cases withuandvintegers Y²

A =u(Xⁿ⁻²−Yⁿ⁻²

A ); X²

A =u(−Xⁿ⁻²

A +Y^p−2) Yⁿ⁻²

A =v(X²+Y²

A ); Xⁿ⁻²

A =v(X² A +Y²) Or

uY²

A =Xⁿ⁻²−Yⁿ⁻²

A ; uX²

A =−Xⁿ⁻²

A +Yⁿ⁻² vYⁿ⁻²

A =X²+ Y²

A ; vXⁿ⁻² A = X²

A +Y²

Or Y²

A =u(Xⁿ⁻²−Yⁿ⁻²

A ); X²

A =u(−Xⁿ⁻²

A +Yⁿ⁻²) vYⁿ⁻²

A =X²+ Y²

A ; vXⁿ⁻² A = X²

A +Y² Or

uY²

A =Xⁿ⁻²−Yⁿ⁻²

A ; uX²

A =−Xⁿ⁻²

A +Yⁿ⁻² Yⁿ⁻²

A =v(X²+Y²

A ); Xⁿ⁻³

A =v(X² A +Y²) First case

Yⁿ=uv(A²Xⁿ−Yⁿ+A(Y²Xⁿ⁻²−Yⁿ⁻²X²))

=uv(A²Xⁿ−Yⁿ+A(AZⁿ)) =uv(A²Xⁿ+A²Aⁿ−Yⁿ) =uv(A²Yⁿ−Yⁿ) Thus

uv= 1 A²−1

Asuvis integer, it means that it is impossible thusu= 0andA²= 1orA=±−1 (Ais an integer and can not equal to√

2) it means thatq= 3andp= 2.

Second case

uvYⁿ

A² =Xⁿ−Yⁿ

A² +Y²Xⁿ⁻²−Yⁿ⁻²X² A

=Xⁿ−Yⁿ

A² +Zⁿ=Xⁿ+Zⁿ−Yⁿ

A² = (A²−1 A² )Yⁿ Thus

uv=A²−1 And

uv(Y²Xⁿ⁻²−X²Yⁿ⁻²) =uvAZⁿ =u(X²ⁿ⁻⁴−Y²ⁿ⁻⁴)A=v(X⁴−Y⁴)A

Thus

uZⁿ=X⁴−Y⁴; vZⁿ=X²ⁿ⁻⁴−Y²ⁿ⁻⁴ uv=A²−1 = (X⁴−Y⁴)(X²ⁿ⁻⁴−Y²ⁿ⁻⁴)

= (Y²Xⁿ⁻²−X²Yⁿ⁻²)²−1 =X²ⁿ+Y²ⁿ−Y⁴X²ⁿ⁻⁴−X⁴Y²ⁿ⁻⁴

=Y⁴X²ⁿ⁻⁴+X⁴Y²ⁿ⁻⁴−2XⁿYⁿ−1 And

X²ⁿ+Y²ⁿ+ 2XⁿYⁿ= 2Y⁴X²ⁿ⁻⁴+ 2Y²ⁿ⁻⁴X⁴−1

= (Yⁿ+Xⁿ)² = (2Yⁿ−Zⁿ)² = 4Y²ⁿ−4ZⁿYⁿ+Z²ⁿ Ifn≥3then

Z²ⁿ+ 1

Y = 2Y³X²ⁿ⁻⁴+Y²ⁿ⁻⁵X⁴−2Y²ⁿ⁻¹+ 2Yⁿ⁻¹∈Z And It is impossible ! It means thatn= 2.

Third case : We have here

Y²=u(AXⁿ⁻²−Yⁿ⁻²); X² =u(−Xⁿ⁻²+AYⁿ⁻²)

vYⁿ⁻² =AX²+Y²; vXⁿ⁻² =X²+AY² And

vYⁿ=u(A²Xⁿ−Yⁿ+A²Zⁿ) =u(A²−1)Yⁿ v =u(A²−1)

v(Y²Xⁿ⁻²−X²Y^p−2) =vA=uvA(X²ⁿ⁻⁴−Y²ⁿ⁻⁴) =A(X⁴−Y⁴)

=u²A(X²ⁿ⁻⁴−Y²ⁿ⁻⁴)² =v²A Thus

v= 1 =u(A²−1)

WithuandA²−1integers, it meansA²= 2: Impossible ! Fourth case : uY²

A =Xⁿ⁻²−Yⁿ⁻²

A ; uX²

A =−Xⁿ⁻²

A +Yⁿ⁻² Yⁿ⁻²

A =v(X²+Y²

A ); Xⁿ⁻²

A =v(X² A +Y²) We have here

uY²=AXⁿ⁻²−Yⁿ⁻²; uX²−AYⁿ⁻² =−Xⁿ⁻² And

Yⁿ⁻² =AXⁿ⁻²−uY² = (Y²Xⁿ⁻²−X²Yⁿ⁻²)Xⁿ⁻²−uY² Hence

uYⁿ

A² =v(Xⁿ−Y^p

A² +Zⁿ) =v(1− 1 A²)Yⁿ Thus

u=v(A²−1)

u(Y²Xⁿ⁻²−X²Yⁿ⁻²) =uA=A(X²ⁿ⁻⁴−Y²ⁿ⁻⁴) =uv(X⁴−Y⁴)A u=X²ⁿ⁻⁴−Y²ⁿ⁻⁴ =v(X⁴−Y⁴) =uv(X⁴−Y⁴)

Thusu = 1andv(A²−1) = 1withvandA²−1integers, it meansA²−1 = 2: Impossible !

The only solution, in all cases, inn= 2.

And there is of course the trivailn= 1

Conclusion

Fermat equationYⁿ =Xⁿ+Zⁿhas solutions only forn = 2. We have shown a way to solve it.

Références

[1] Paolo Ribenboïm, The Catalan’s conjectureAcademic press , (1994).

[2] Robert Tijdeman, On the equation of CatalanActa Arith , (1976).