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September 13th, 2015, 06:44 PM   #21
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Quote:
Originally Posted by CRGreathouse View Post
Are you moving the goalposts on me?
I'm not. Sorry, if you take it this way.
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September 13th, 2015, 08:25 PM   #22
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Quote:
Originally Posted by victorsorokin View Post
Fermat's Last Theorem (main case: n is prime):
For integers A, B, C and prime n>2 the equal $\displaystyle A^n+B^n=C^n$ does not exist.

The essence of the contradiction: If Fermat’s equality $\displaystyle A^n+B^n=C^n$ exists then A+B-C=0 and $\displaystyle A^n+B^n<C^n$.
False. If $(i)\ A^n+B^n=C^n$ is true, in no way it follows that $(ii)\ A+B-C=0$, only the contrary trivially holds. One can show that $(i')\ A^p+B^p=C^p$ implies $(ii')\ A+B-C\equiv 0\pmod p$, but that's a whole different thing.

These sort of elementary logical flaws mixing necessary and sufficient conditions, and confusion between equality and equivalence, thrive in all so-called FLT Proofs I see in this forum, I wonder whether they come from the same source...
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September 13th, 2015, 11:03 PM   #23
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Well, that's one actual critique so far. Here's
another:

"Fermat's Last Theorem (main case: n is
prime): For integers A, B, C and prime n>2 the
equal $\displaystyle A^n+B^n=C^n$ does not
exist."

That's fine except n = 2 is also prime and I can
not see where the argument accounts for this
case. [See my next note, below.]

"The essence of the contradiction: If Fermat’s
equality $\displaystyle A^n+B^n=C^n$ exists.."

I interpret this to mean that if there are any
integer solutions to A^n + B^n = C^n for n > 2
then...

"... then A+B-C=0 and $\displaystyle
A^n+B^n<C^n$."

A^n + B ^n < C^n is true for any n > 1 if A + B
= C; so how exactly does the argument not
contradict the case n = 2?

Further in the paper:

"So, let us assume that for a prime number
n>2, relatively prime A, B, C, and A'[or B']≠0
1°) $\displaystyle A^n=(C-B)P$,..."

This seems to assume the number A is always
the product of two different factors. If that is
the assumption then no - there is no reason
why A can not be prime.

That's the best I can do because it's the most
I can almost (!) understand.
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September 14th, 2015, 12:23 AM   #24
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I felt many time in similar errors. The most difficoult problem in FLT is to arrive at a conclusion that cannot live space to round cicles...(or classic stupid errors).

I was finally arrived ad the concernig that the only way is to use a more simple modular algebra (more than the one udes by Wiles) and prove the point that is the non vanishing mixed product from n>2 that cause the proof of flt.

I invent the complicatemodulus algebra, but I turn around for several years since I'm not able to clearly prove what I've in mind and what seems to me "logic". Logic is not a proof... a reduction as absurdum is a proof, or a counterexample.

...And is what I hope I found this morning starting from:

$\displaystyle (C^n/B^n - A^n/B^n) = 1$

using my step sum in the proper manner

Going to the limit and proving that in the case n>2 the mixed product seems doesen't vanish living us another equation that contracdict the (1).

Of course for n=2 the absence of the mixed product let the solution be possible.

I work on and in the next week I let you know...

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Ciao
Stefano
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September 14th, 2015, 01:58 AM   #25
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Quote:
Originally Posted by complicatemodulus View Post

...Going to the limit .....
Pls turn in:

"the solution can be found JUST going to the limit, so variable in R (not in Q, not in N) when the mixed product terms will vanish..."

n=2 works since there are just 2 terms so no "mixed" terms (equal to flat derivate).

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Stefano
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September 14th, 2015, 02:23 AM   #26
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Quote:
Originally Posted by complicatemodulus View Post
I invent the complicatemodulus algebra, but I turn around for several years since I'm not able to clearly prove what I've in mind and what seems to me "logic". Logic is not a proof... a reduction as absurdum is a proof, or a counterexample.
RAA (reductio ad absurdum) is part of classical logic, and for these sort of proofs basic propositional logic would help a lot. Draw a truth-table for the material conditional (implication).

You will easily see that when $P\Rightarrow Q$ is true, it also can be the case that $P$ is false, so many people commit fallacies such as: since the $implication$ is true, so is the antecedent $P$, what is false. Or: since the $implication$ is true, so is the antecedent $P$ and the consequent $Q$, what is also false. Or: since the $implication$ is true and the consequent $Q$ is also true, so is the antecedent $P$, FALSE.

Now lets see a silly example: $P=$"somebody is alive" and $Q=$"somebody was a baby". Now, it is sufficient to say that somebody is alive to conclude that somebody was a baby in the past. It is necessary to have been a baby in order to be alive (we still don't engineer complete adults in genetic labs). Then $P\Rightarrow Q$ is a true implication.

But suppose that you know someone deceased. $P\Rightarrow Q$ and $Q$ is also true of that person. It does NOT follow that the person is alive.

Not, put $P=''A+B-C=0''$ and $Q=''A^n+B^n-C^n=0''$, then assuming that $A^n+B^n-C^n=0$ is true, then $P\Rightarrow Q$ is emptly true (it may be the case that $A+B-C\not=0$ but that wont matter). But if we say that $Q\Rightarrow P$, we are saying that $P=A+B-C=0$ is a necessary condition for $Q=A^n+B^n-C^n=0$ to be true, the same way that "somebody was a baby" is necessary to infer that "somebody is alive", what is plainly false.

Last edited by al-mahed; September 14th, 2015 at 02:25 AM.
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September 15th, 2015, 05:06 AM   #27
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magazines

Quote:
Originally Posted by CRGreathouse View Post
Really, journals don't care who you are, they just care about what you produce. If you write good things and they publish them, their prestige increases. If you write great things and they reject you, they have egg on their face.
Often magazines perceive little typos as large error and stop communication. In this respect, the forum is not such a tough opponent.
=
I sent the proof to Polish mathematical journal.
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September 15th, 2015, 05:19 AM   #28
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contradiction

Quote:
Originally Posted by al-mahed View Post
False. If $(i)\ A^n+B^n=C^n$ is true, in no way it follows that $(ii)\ A+B-C=0$, only the contrary trivially holds. One can show that $(i')\ A^p+B^p=C^p$ implies $(ii')\ A+B-C\equiv 0\pmod p$, but that's a whole different thing.
You're right, but I'm also right: A + B-C = 0. The contradiction in this is.
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September 15th, 2015, 05:38 AM   #29
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n=2

Quote:
Originally Posted by uvkajed View Post
1) A^n + B ^n < C^n is true for any n > 1 if A + B
= C; so how exactly does the argument not
contradict the case n = 2?


2) This seems to assume the number A is always
the product of two different factors. If that is
the assumption then no - there is no reason
why A can not be prime.
1) If n=2, than a_[2]=an+1 and q_[2]=-an+1 are not equal and A_{2}≠(an+1)^2 !!!

2) Yes.
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September 15th, 2015, 07:35 AM   #30
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Quote:
Originally Posted by victorsorokin View Post
Often magazines perceive little typos as large error and stop communication.
Not really. If the paper is decent they won't care about typos. In my experience they're very understanding.
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