# Thread: Fermat's Theorem, Revistited

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

Originally Posted by BuleriaChk
What calculations does it not account for? grapes ridiculous "c<(a+b)"? (not even relevant for n > 2)
There it is. Not only have you not proven it for c<a+b, you don't even think it is relevant.

How about c=12, a=9, b=10? Have you proved that ? No, you have not, because 12<9+10, and you think it's irrelevant. You couldn't be more wrong.

Of course it's easy to show that unequality -- but you have not done so. That's why your proof fails

A proof has to address all possible values
A quote about Euler's conjecture? What does that have to do with my proof? Or the relativistic unit circle?

(And none of these guys reference independent variables....)

What flaws? (quote my pdf)... I call bullshit... Shut the fuck up if you can't be specific.

You have no idea what you are talking about. You've only posted drivel.

Well, by now the whole world (or at least the few that are reading this thread) know what we think of each other...

Time will tell.... and you're on record...

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

Originally Posted by BuleriaChk
(snip NF) if you can't be specific.
I was very specific. Your "proof" fails to meet the rigorous standard of proving that the claim is consistently and always valid.
(snip NF)
(snip NF)

If you want to discuss your ideas, Chuck... you need to be willing to accept critiquing. Rather than using criticism to better your model, you just change tactics. Rather than correcting errors, you dismiss them. And when all that fails, you go to insults and vulgarity.

Originally Posted by grapes
There it is. Not only have you not proven it for c<a+b, you don't even think it is relevant.

How about c=12, a=9, b=10? Have you proved that ? No, you have not, because 12<9+10, and you think it's irrelevant. You couldn't be more wrong.

Of course it's easy to show that unequality -- but you have not done so. That's why your proof fails

A proof has to address all possible values
Chuck got so caught up in thinking that if 2<n, when he saw the less than sign, c<a+b, he assumed c must be less than 2. It's these simple assumptions and errors that he bases his entire case on and why it falls down like a house of cards no matter how much he tries to snatch them up and place them upright again.

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

Originally Posted by BuleriaChk
My proof is based on the concept that there is only one unique set of integers on a single number line, represented by (a). For generality two relate two different integers, two number lines are needed (a,b) with the relationship given ultimately by the relativistic unit circle.

This means that the only integers that are countable if 1 is chosen as initial state are those for which in a single dimension.

for the final state (1) invariant.

If the initial state = final state, then ,

if the final state is greater than the initial state, then

c= a+b means that the symbols "c" and "a+b" refer to the same unique integer on a single number line , means that c = d, so that (no b required).

You keep repeating the same misconception about the nature of dimensions (not to mention trigonometry, calculus, polynomials, all of which require more than one dimension. Including the equation .

Sheesh, have you EVER graphed a function? Even in high school? Any trigonometry even?

(You might learn something if you read the rest of the proof....)

Village idiot.
again, you cannot assume that c=a+b as we know that c<a+b. You cannot get around it so using binomial expansion is a pointless attempt.

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

In order to clean up this thread, removing whole posts to the Cage Match would have made the thread very difficult to follow. Instead, material was edited out of the offending posts while the On Topic content remains and the vulgarity, ad homs and insults were moved to here.

If you wish to continue the personal attacks in that thread, we can. In this thread, we should try to stay on topic without them.

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

Apparently, there are two of these threads with the same title.

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