Re: Unexpected shot for FLT !

Roman B. Binder a écrit :
Denis Feldman schrieb : (* * *)
Roman B. Binder a écrit :
tbonepower07@xxxxxxxxx wrote:
Roman B. Binder wrote:
matt271829-news@xxxxxxxxxxx wrote:
Roman B. Binder wrote:
Hi,
You are afraid to say something
because of Your boss ?
" Silence tells me everything "
Ro-bin
I think your posts are generally very difficult
to
understand. This
might discourage people from trying to work
through
them.

OK: I'll dismiss not necessary words and some
very well known conditions:
1)-st of my steps for FLT:
X^n + Y^n = Z^n ??? for prime n>=3; X;Y;Z
integers;
then there are such natural x;y;z of gcd=1 that:
Ls = x^n + y^n = z^n = Rs
Lets look for such natural c that:
Ls + c^n = (x^n + y^n + c^n)/xy
c = kxy -(x+y)
once Ls + c^n = (x+c)Ext + y^n = (y+c)Ext + x^n
Ls+c^n =(x+kxy-x-y)Ext +y^n =(y+kxy-x-y)Ext
+x^n
Ls+c^n =y(kx -1)Ext + y^n = x(ky -1)Ext + x^n
Ls+c^n = y[(kx-1)Ext + y^(n-1)] =
= x[(ky-1)Ext + x^(n-1)}
Ls+c^n = (x^n + y^n + c^n)/ xy
for c = kxy -x -y
( generally true also for c;k integers
and for all n positive odd numbers
so for sure for n>=3 and prime numbers
and for k natural numbers )
is it clear till now ???
This is first step introduced also as
Lemma: let X^n + Y^n + C^n = L for odd n >0
then for C = KXY -(X+Y) and X;Y;K integers
L/XY
More two my steps and FLT is ours in much more
easy way: if it is difficult for You to
understand
so may be it is not so elementary way:
2) In the second step we'll try to find such
simple z that z^n + our c^n will be also
divisible
by xy
3) In the third step we'll try to generalize all
possible cases of factorization of z^n + c^n
for to achieve (z^n + c^n)/xy
Again,is it clear till now ?
Can I continue my step 2 ?

With Compliments
Ro-bin
Unless I misunderstood, you said

Ls + c^n = (x^n + y^n + c^n)/xy

but you also said

Ls = x^n + y^n

so you are saying

x^n + y^n+c^n = (x^n + y^n + c^n)/xy
1= 1/xy
If x and y are integers,
x = 1 and y = 1

So that's a flaw right there.
obviously not:
* * *
You are probably trolling, but in the offchance you
are sincere, you
I am not trolling but You used to sleep during my
lesson:


I am not your pupil (and not as asleep as you believe), and if you want to communicate, it could be a good idea to use common conventions (and if you dont know them, ask, butsuspect your ideas will not be taken very seriously)


I can call Ls or L and such symbols only
one time be appropriate for left side of certain equation;

Extremely bad practice. Much better would be let L_1=x^n+y^n+c^n ; and later ; let now L_2 be ...


In my first topical
I used Ls = x^n + y^n + c^n and I used to find certain c values:
c = kxy -(x+y) , that Ls/xy or You like
the use of xy/Ls so I am sorry and I'll
try to edit my topics this way.
Is it all right now ?!


Certinly not. Stop using a/b for "a divides b" (a|b is better) , but more important, stop writing things like "1+1=2/4". You are not in such a need of abreviations. At least, use "1+1=2 and 2|4".

Next problem is motivation. Why is c of this shape?

Next is this paragraph :

once Ls + c^n = (x+c)Ext + y^n = (y+c)Ext + x^n

What is Ext? Certzainly not the qame on both sides of the = sign...


Why is this true (and what means once?)



>>>>> Ls+c^n =(x+kxy-x-y)Ext +y^n =(y+kxy-x-y)Ext
>> +x^n
>>>>> Ls+c^n =y(kx -1)Ext + y^n = x(ky -1)Ext + x^n
>>>>> Ls+c^n = y[(kx-1)Ext + y^(n-1)] =
>>>>> = x[(ky-1)Ext + x^(n-1)}
>>>>> Ls+c^n = (x^n + y^n + c^n)/ xy
>>>>> for c = kxy -x -y
>>>>> ( generally true also for c;k integers


must realize the proble stems from your use highly
non standard of /, your sloppy use of =, and a few other mism atches in
notation. Read the answer again, and realize what was the problem, then
try to write your first line (and the others) according to this model:

Lets look for such natural c that:
Ls + c^n = (x^n + y^n + c^n) is divisible by
y xy



Good luck
* * *
Thank You ( I am not running from colours yet,
but I can use my signs, for this matter will
be readable also for other readers )
for c = xy -(x+y); x=1;y=1
c = 1 - 2 = -1
1^n + 1^n -1^n = 1^n divisible by 1 =xy
Initially Lemma was indicated for x;y natural
numbers of gcd=1, like these resisting in FLT
question. I have seen chance for to retrieve more
general results, but for optional integers I can
now check some special configurations:
1) x=-y L=x^n +y^n +c^n
for c = kxy -(x+y)
L= x^n -x^n +(kxy -x+x)^n = (kxy)^n /xy
I can also return to general forms:
L = x^n +y^n +(kxy-x-y)^n configures to
x[(ky-1)Ext +x^(n-1)] = y[(kx-1)Ext +y^(n-1)]
so anyhow L/xy May be You'll find better counter example ?

I will certainly not look for them. Either you give proof of your results, or at last indication of plausibility, or else nobody will even try to look at them.






Compliments
Ro-bin
.

Relevant Pages

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    ... I think your posts are generally very difficult ... for c = kxy -x -y ... More two my steps and FLT is ours in much more ... You are probably trolling, but in the offchance you are sincere, you must realize the proble stems from your use highly non standard of /, ...
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  • Re: Unexpected shot for FLT !
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    (sci.math)
  • Re: Unexpected shot for FLT !
    ... might discourage people from trying to work ... 1)-st of my steps for FLT: ... for c = kxy -x -y ... You are probably trolling, ...
    (sci.math)
  • Re: Unexpected shot for FLT !
    ... " Silence tells me everything " ... I think your posts are generally very difficult ... for c = kxy -x -y ... More two my steps and FLT is ours in much more ...
    (sci.math)
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