high everybody (nobody included)

I need your help to find the mistake probably hidden in this very short attempt of a demonstration of the LFT théorème (please excuse my english that is not so good as it should be ( I will put "(?) when I am not sure if I use a frenchissim(?)))

So lets go

lema 1 : if A and B are prime together and B is prime whith p (where p>2 is prime ) for any integer N we can find L > N for wich A+LB= p' is prime and mod(p,p') =2 (I use Exel notation

wich i supose are international(?))

I will not demonstrate(?) this elementary corolairy(?) of Dirichelet Théorem (thanks to him) for I suppose the "helper" are able to do that themselves

Lema 2 : if p is prime ,p' prime and not(mod(p,p')=1), A and B prime together

we have (A^p +B^p )// p' => (A+B)//p' (a//b means a and b are not prime together which is my way to say there GCD(? greater comon divisor ) is not one)

This is a not as so evident (as upstair) colorairy of Fermat Théorem (the "litle" one (this is not a recursive demonstration).

So if as me you agree whith those lema here come the suspect stuff:

we supose we have got p>2 prime X ,Y (not Y//p), Z , ((X,Y,Z) mutualy prime together(?))

whith X^p + Y^p = Z^p and we are in search of this => an impossible thing

we consider the number A built using the folowing receipe(?) :

for each (p'') prime number between 2 and (2 * Y)+1 wich is not a divisor of Y we search (and find) Ap'' which is the smallest number whith the form p"^k which is greater than (2 * Y)+1

A is then the square of the product of all those Ap" (I know that i should have take a beter

notation I hope you have understand whath i meant to say ( i take a great margin by taking the square to avoid colateral damage)

Then we consider (AZ)^p +L(Y)^p=p' whith p' prime and mod(p,p')=2 provided by lema 1

So as i am not very consize I will continue this stuff in an own reply

so we have got (AZ)^p+L(Y^p)= p' and mod (p,p')=2 (this is a reminder)

Then we use (and you probably know) that we can find B<p' whith mod ((B^p - A^p),p')=L

so we have (AZ)^p +(B^p -A^p)*Y^p = (AX)^p + (BY)^p (by hypothesis) // p'

here is the critical point

whe note C' the GCD (see above) of A and B ,A'=A/C' ,B'=B/C'

so we have according to lema 2 : (A'X)+B'Y=Cp' where C is obviosly < 2Y+1

supose that not(C=1) then there is p">1 and k' whith p''^k' divide C and p''^(k'+1) do not

(I dont know wy i am so laxatif(?) you perfectly know where i want to go)(and where i canot go this way i think because i made a mess i dont thing i can proove this way that A' is divisable(?) by p"^k whith k'<=k which would make B=(Cp'-AX)/Y // p" so C' could not be the GCD of A and B

nevertheless I folow my idea ( i should had a lema 3 maybe)

the point is to find A'X+B'Y =p'

an A''X+B''Y=p' where if C'' is the GCD of A and B+p' , C"B"=C'B'+p' = B+p'

It would be then easy to conclude considering we are using positive integer and that L (see above) is as biger as we would need

thanks for your patience(?)

By By p.p.