Thread: not so very short demo of LFT ?

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.

...Down in the Mississipi
In a fload...
John Cale (InTheFload-Black Acetate)

So all of your twenty-four readers, You don't seems in a hurry to contradict me.
I wasn't expecting too many anyway!
I made a few mistake but they don't seems to be lethal to the general dublin of my demo:
In lema 2 you don't need the two positives integers to be mutualy prime to get the implication.
In the "delicat point" as A is built to contains any prime divisor betwen 2 to 2*Y+1 prime(?) whith any divisor of Y, C' or C (I don't remember) must be equal to 1 because if not B'=(C'p''-AX)/y would be divisable by a divisor of A say p''' the min of the non equal to one divisor of C wich is not divisor of Y (because AX is prime whith Y).
So you have not awnsered my question yet.
It is important because if the demo is Holding On (as you my dears) I will put it the correct way (ie in Peano's language) which is a hard job.
Have a Wonderfull Day in our's "Game Whithout Frontiers" world

P.Picaziet
Directeur de la banque d'en face.
Membre de l'académie de pataphysique.
Chercheur en emploi (avec le Code du Travail en main)
Usufruitier du tas de repassage qui m'attend (salut)
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One potato two potatoes tree potatoes four
Five potatoes six potatoes seven potatoes more..
traditional Irish

i don't understand a foxtroting word of what you had writed (but i am not the only one)
The only free lunch théorem that i know is : there is no free lunch
(exercise proof it (if it's true)).
anyway why are you tring to make things simple when we can make it complicate?
may be someone will dare to give you his opinion on the mater but they don't seem to have this sort of courage.
don't feel isolated i am whith you.
Cheers
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6 by 6 from wall to wall
the shuders on the windows, no light at all
...ect...ect
P.G.