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May 16th, 2017, 01:34 AM   #21
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Is more far as you think since the most simple proff comes just looking to the numbers of what I've posted above so to the inequality:

So we prove we can give a solution to the inequality:

\[ Y_P' > Y'_{MP} > Y'_{BP} \] so to:

\[ Y_P' > Y'_{MP} \] so to:

\[ [2n(X_f)^{(1/n)}]^{(n-1)} > [(2nA^{(n-1)}+((nB^{(n-1)}-nA^{(n-1)}))] \] or:


\[ 2n[(A^n+B^n)^{(1/n)}]^{(n-1)} > [(2nA^{(n-1)}+((nB^{(n-1)}-nA^{(n-1)}))] \]


That luckily for me is well sorted, so onve we prove (by comutation) the case $n=3$ with $A=5$, $B=6$ any other Bigger combination will return RISING DIFFERENCE, so we can be sure there are no zeros somewhere in the middle, just using the comparison criterion, and knowing that, viceversa going infimus in the other direction we always get a solution when $A=B$.







Infact if we try to say somethink on $X$, we can't say somethink since the value of $X_f$ can be Bigger or Lower $X_MP$ so we have in case to prove:

\[ X_f \neq X_{BP} \] so to:

\[((B^n+A^n)/2]^{(1/n)} \neq [(A^{(n-1)}+((B^{(n-1)}-A^{(n-1)})/2)]^{(1/{(n-1)}} \]




That again fells in knowing FLT solution by Wiles proof...

I'm still working in a good definition of all the points... will be a long work... but now the main "subject" of the proof comes out...

I know that from a long time by numbers, but to prove in "math" was a veery long way... and I admit just question of luckyness to find a well sorted Ordinate...

The proof involves Ordinal Numbers can holds too, but will be very hard for all is not aware of that concepts.

Last edited by complicatemodulus; May 16th, 2017 at 01:50 AM.
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May 17th, 2017, 10:40 AM   #22
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Originally Posted by Mariga View Post
Is there anything wrong with the proof?
Yes. When n=2 assumption that (c+a) and (c-a) have common divisor k means it divides each term c, a, and b and must be reduced.
When n>2 the only possible common divisor of (c-a) and [c^(n-1)+c^(n-2)b+... +b^(n-1)] is n. Division of latter by former gives remainder na^(n-1). So they cannot have common divisor k.
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May 18th, 2017, 04:55 AM   #23
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Originally Posted by McPogor View Post
Yes. When n=2 assumption that (c+a) and (c-a) have common divisor k means it divides each term c, a, and b and must be reduced.
When n>2 the only possible common divisor of (c-a) and [c^(n-1)+c^(n-2)b+... +b^(n-1)] is n. Division of latter by former gives remainder na^(n-1). So they cannot have common divisor k.
No.. it doesn't mean that the divisor divides each term c, a and b. Take any primitive pythagorean triples as an example. I will take 3,4,5. c=5 and a=3 or 4 and b= 3 or 4.

If a=3, c-a=2 c+a=8. In this case, k=2, x1^2=1, x2^2=4

If a=4, c-a=1 c+a=9. so k=1, x1^2=1, x2^2=9
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May 18th, 2017, 05:54 AM   #24
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It is incorrect to suggest that $b^n$ must be a product of integers $(kx_1^n)(kx_2^n)(kx_3^n)\cdots (kx_n^n)$. Or rather, there are other ways to form $b^n$. For example $$b^3=(x_1^3)(kx_2^3)(k^2x_3^3)$$
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May 18th, 2017, 05:57 AM   #25
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The suggestion that Fermat's idea for the general case was the same as his proof for $n=4$ doesn't follow. If his general case had worked, he'd never have needed a proof for $n=4$.
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May 18th, 2017, 09:59 PM   #26
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Originally Posted by v8archie View Post
It is incorrect to suggest that $b^n$ must be a product of integers $(kx_1^n)(kx_2^n)(kx_3^n)\cdots (kx_n^n)$. Or rather, there are other ways to form $b^n$. For example $$b^3=(x_1^3)(kx_2^3)(k^2x_3^3)$$
Indeed.But it is still the same thing. We can divide $k^2x_3^3$ by $k$ and multiply it by $x_1^3$ to get $(kx_1^3)(kx_2^3)(kx_3^3)$
same as $k^3x_1^3x_2^3x_3^3$ for the two cases.

Also still in both cases, $b^3$ is a product of integers that can be expressed in three groups. Hence we have to show $c^3-a^3$ is not a product of any 3 combinations/operations of $c$ and $a$ such that the combinations are integers.

Though the proof has to be corrected and that point noted down.

Last edited by Mariga; May 18th, 2017 at 10:19 PM.
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May 18th, 2017, 10:14 PM   #27
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The suggestion that Fermat's idea for the general case was the same as his proof for $n=4$ doesn't follow. If his general case had worked, he'd never have needed a proof for $n=4$.
He might have followed a procedure related to the case of $n=4$. I guess it would have been lengthy. We just wonder why he never wrote it down yet the only limiting factor would have been time.. not the margin. Anyway he will forever remain to be the only person who knew whether he really had the proof. He could have his own reasons for not sharing.
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May 18th, 2017, 10:22 PM   #28
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Anyway he will forever remain to be the only person who knew whether he really had the proof. He could have his own reasons for not sharing.


There's a false proof that would work if only some ring or other were a unique factorization domain but it isn't. Some people think Fermat THOUGHT he had a proof but actually didn't.

On the other hand he never mentioned the matter again over the next 30 years. It really is unlikely he had a proof and he probably did know he didn't have it.
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May 19th, 2017, 05:03 AM   #29
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Quote:
Originally Posted by Maschke View Post
There's a false proof that would work if only some ring or other were a unique factorization domain but it isn't. Some people think Fermat THOUGHT he had a proof but actually didn't.

On the other hand he never mentioned the matter again over the next 30 years. It really is unlikely he had a proof and he probably did know he didn't have it.
Or he thought he had a proof when he scribbled his assertion on a margin. but when he tried to develop it on something larger than a margin, it disappeared. Standards of proof in the 17th century were not as rigorous as they have since become, but the 17th century certainly had standards of proof.
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May 25th, 2017, 07:36 AM   #30
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Take a look at this
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