1. ## Re: Fermat's Theorem, revisited

Buleria, are you really this delusional?

2. ## Re: Fermat's Theorem, revisited

Here is the real solution to Fermat's Theorem (see my ignore list)

The Binomial Expansion is of the form (a + b)n = an + bn + rem(a,b) where rem (a,b) all powers of a and b and coefficients are positive integers. Fermat's Theorem is:

an + bn cn for n > 2, a and b positive integers

In order for Fermat's Theorem to be false (i.e., an + bn = cn) for some n > 2, a and b positive integers, rem(a,b) must be equal to zero, which it is not, by inspection (since rem(a,b) > 0, with a and b and the coefficients are positive integers for all powers of a and b and all n > 2.

(Edit: I had a senior moment where I wrote "true" a short while ago) I have written it correctly numerous times in other threads. If this is what Grapes is referring to, I can only say "my bad"; sorry, Grapes if you couldn't figure that out from the context.) .. (There is a whole other thread where I say it correctly, but WAY over engineer the problem - I simply couldn't believe it was that easy. I'm starting to believe the whole thing is a huge joke, like believing Trump can be president (just ask Vince McMahon if it is just another WWE passion play). Well, ok, if it weren't so serious...

===============================
For the case n = 2 (note: Fermat's Theorem is not relevant here, so this is actually a fool's errand w.r.t. to the hypothesis),

we must have

c2 - (2ab) = a2 + b2 is a positive integer

Therefore for Fermat's theorem to be true for this case, we must have a = 0 or b = 0, so the equation is not satisfied.
(there must be an (integer) metric for the integers as a ring without division, so 0 is the midpoint of all possible integer "distances" (lengths) defined in relation to 0 (a - a) = (b - b) = 0 for all a and b.)

QED

Try calling the Abel committee again... they may be finished their conversation with John....

3. ## Re: Fermat's Theorem, revisited

Originally Posted by emperorzelos
Buleria, are you really this delusional?
Yep

But it's fascinating
Originally Posted by BuleriaChk
Here is the real solution to Fermat's Theorem (see my ignore list)

The Binomial Expansion is of the form (a + b)n = an + bn + rem(a,b) where rem (a,b) all powers of a and b and coefficients are positive integers.

In order for Fermat's Theorem to be true:

an + bn cn for n > 2, a and b positive integers,

rem(a,b) must be equal to zero, which it is not, by inspection (rem(a,b) > 0 since a and b are positive integers.
rem(a,b) is never zero, not even for n=2

But Fermat's Last Theorem is not about (a+b)n equaling an+bn. We know that c < (a+b)
===============================
For the case n = 2,

we must have

c2 - (2ab) = a2 + b2 is a positive integer

The Fermat condition requires a = b = c = 0 (since 2ab = 0), so the Binomial Theorem (and Fermat's Theorem) is not relevant for this case...
No, c cannot equal a+b

You've made that error over and over
QED

Try calling the Abel committee again... they may be finished their conversation with John....
The Abel committee would insist on a GED, at least

4. ## Re: Fermat's Theorem, revisited

Originally Posted by grapes
Yep

But it's fascinating

rem(a,b) is never zero, not even for n=2
That is the whole point of my proof.

Think about it and try harder to understand Fermat's Theorem.
And the Binomial Expansion (you'll need it)

Originally Posted by grapes

But Fermat's Last Theorem is not about (a+b)n equaling an+bn. We know that c < (a+b)

No, c cannot equal a+b

You've made that error over and over

The Abel committee would insist on a GED, at least
Please, Grapes, try - I mean REALLY TRY to understand Fermat's theorem....

(I never said c = a + b) (i.e. c1 = a1 + b1) That statement is not even relevant (since n 1 in Fermat's equation to begin with)

5. ## Re: Fermat's Theorem, revisited

It is interesting that Fermat was born some time after the Binomial Theorem was introduced.

I'm thinking he just put 2 and 2 together for his theorem, but didn't have the space for the Binomial Theorem in the margins of his paper.....

6. ## Re: Fermat's Theorem, revisited

Originally Posted by BuleriaChk
(I never said c = a + b) (i.e. c1 = a1 + b1) That statement is not even relevant (since n 1 in Fermat's equation to begin with)
You said , which is the same as c = a+b, when you're dealting with positive integers

Elementary algebra, it's what I've been pointing out for months, and you ignore it.

7. ## Re: Fermat's Theorem, revisited

Originally Posted by grapes
You said , which is the same as c = a+b, when you're dealting with positive integers

Elementary algebra, it's what I've been pointing out for months, and you ignore it.
Try some numbers; e.g. a = 3, b = 4, c = 7

c2 = (a + b)2 = a2 + 2ab + b2 = 9 + 24 + 16 = 49

( (a+b) = c = 7)

Hey, but it is equal to c2 = 72

MONTHS?..

8. ## Re: Fermat's Theorem, revisited

Most of my other posts were because I simply could not believe it is that easy. It may not be, but if I'm wrong it is not for the reasons those on my Ignore list claim.... I really do think it is a huge mathematical "in" joke now...

9. ## Re: Fermat's Theorem, revisited

Originally Posted by BuleriaChk
MONTHS?..
Almost a year:

Fermat's Last Theorem

10. ## Re: Fermat's Theorem, revisited

Originally Posted by grapes
You said , which is the same as c = a+b, when you're dealting with positive integers

Elementary algebra, it's what I've been pointing out for months, and you ignore it.
see post #17
And demonstrably wrong that whole time?

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