# Thread: Proof of Fermat's Theorem by Modulus

1. ## Re: Proof of Fermat's Theorem by Modulus

Originally Posted by BuleriaChk
You say the "proof" isn't valid, but have no fricken idea why it is not. It is a polite way of braying "nonsense" like a polite donkey, but nevertheless a donkey just the same.
I have pointed out the flaws and so has Grapes, and we both have pointed out the same flaws more times than I can count.
You braying "It's not flawed" is no different. You are wearing blinders. Grapes addresses the Same Flaws he has been pointing out since Page One of your Fermats Last Theorem thread not even five posts ago!
Yet, you still are in complete denial.
You always claim that I didn't post a link or didn't support the statement or whatever, utterly ignoring the links in black and white on the page. You are so notorious for it that when I post a Link, I even dare you to forget its existence.
So, it is no surprise you use that same tactic here and now. You are often consistent in your trappings.
Originally Posted by BuleriaChk
And the other statement is without content unless you actually know how to resolve the paradoxes.
NONSENSE!!!
Paradoxes only show that a theory needs work. Or someones understanding of it does... Like the Twin Paradox. In the physical world or reality, no actual paradoxes are ever observed.

Originally Posted by BuleriaChk
Like Grapes, you confuse arithmetic operations with vector operations, and I doubt you understand the latter at all.
Meanwhile it is glaringly plain as day that your assumption does not prove all calculable outcomes of your "Proof."
Originally Posted by BuleriaChk
fricken village idiots....
Are you John Gabriel? You sound like John Gabriel... Is that like Braying "nonsense?"

Originally Posted by BuleriaChk
(like the last comment from Grapes, where c < a+b can be used to create any number relation; e.g. 1 < (1 + 1) and so , and is in no way related to Fermat's equation), and so is irrelevant, whereas actually does mean something, provided you understand and and the relativistic unit circle.
Funny you say that... Your "proof" depends on c=a+b.
Originally Posted by BuleriaChk
And if you have nothing to say besides braying and fan-boy remarks, then as a moderator you should shut the fuck up, instead of spamming threads where technical issues are being discussed.
In spite of your consistent arrogant remarks in which you believe that everyone is stupider than thou... I understand your "proof" from beginning to end and from what I have seen in the posts... so does Grapes and so does MathNerd and so does Emperorzelos... and all have pointed out the same flaws over and over. Part of the problem is how much you clutter it up with circus side-show tangents every time you get caught with an error.
You claim that only arithmetic errors are pointed out... But that is intellectual dishonesty. You and every other reader can plainly see that when you go in and try to fix them, you end up frustrated because you are unable to fix many of the errors without destroying the "proof" you were trying to create.
You then start a new circus side show about Fermi or Pauli or Vectors or what have you... misusing signs, confusing what the word transcendental means (You really got upset over that one...) to distract attention away from the original point that was easily shot down over something trivial and that you could not save.
Your latest tangent is the "Relativistic Unit Circle" and it won't save your 'proof' because your 'proof' is only an assumption; it does not demonstrate that it always satisfies. Which is necessary to be a proof and why Wiles used the modular form. But you don't know what he did... in spite of your frustrated utterances in these threads, you haven't taken the time to examine the accepted working proof that exists in the Real World outside of your delusions of grandeur.

Now...

Telling me to "shut the fuck up" is effectively you wanting to silence me--- maybe because you fear what I might say.
Grapes points out an error or a flaw... So, you call him the Village Idiot. You follow the same trend without fail... When shown your errors you:
-Deny. Fallacy of the Inverse
-Shift the goal posts; run Distraction tangents. Non sequitor; Fallacy of Shifting Ground
-And at the very end, you claim irrelevance on relevant material followed up with a closing "Quitting of the discussion."

2. ## Re: Proof of Fermat's Theorem by Modulus

Originally Posted by Neverfly

Funny you say that... Your "proof" depends on c=a+b.
The proof is obvious from inspection by expanding where rem(a,b,n) > 0 means (for the milliionth time) for n > 2 That this expansion is true for all real numbers (not just integers) is called the Binomial Theorem (Google it).

If you had tried this expansion even once with actual positive integers, you would see it is true. Try a few cases of a,b,n...

Because a and b are indpendent integers, the case is true for all a and b (since nothing changes either of them in the expansion).

THAT is a bried technical explanation, but is proven thoroughly with the relativistic unit circle (which covers all cases of real numbers, like the Binomial Theorem.

(I am exhausted, and it is late at night, so the above analysis may be not quite correct, but if it is not, I have provided many versions of the proof that are. But your comment means you didn't understand any of them, not even the simplest above, that comes from the Binomial Theorem by inspection.)

Anyway, my .pdf's are available and are the resource if you want to understand what I actually said, instead of fantasizing that you understand it or simply lying that you do without any support. I assure you, you do not. If you don't have a technical argument, then just shut up.

Especially as a moderator when the discussion is technical.

(At least Grapes ALMOST understands it (I think) - what he is having emotional problems about is that I may be right, which is why he is desperately throwing shit against the wall to see if anything will stick.) Even to the point of blatantly lying about my results in his own thread which he has now closed.

Grapes is also a fraud as a moderator.

3. ## Re: Proof of Fermat's Theorem by Modulus

As far as Grapes contention that c < a+b is concerned, such an equation means there must be an integer k such that c+k = a+b. Fermat's Theorem contains no such reference to k. Therefore, the Grapes' equation is irrelevant; i.e., k=0 in Fermat's theorem, and the Binomial Expansion is valid in representing the fact that Fermat's theorem is true.

But the proof from the relativistic unit circle and Lorentz rotation proves it for all numbers, not only integers.

4. ## Re: Proof of Fermat's Theorem by Modulus

Originally Posted by BuleriaChk
As far as Grapes contention that c < a+b is concerned, such an equation means there must be an integer k such that
c+k = a+b. Fermat's Theorem contains no such reference to k. Therefore, the Grapes' equation is irrelevant.
No, it is a natural consequence of how the equation works.

If you have then out of neccesity . This is a trivial thing to prove so you cannot ever set c=a+b for it to work in fermats last theorem. The moment you set the equality, you are no longer dealing with fermals last theorem as we know that the equality must be false or it to relate to FLT.

5. ## Re: Proof of Fermat's Theorem by Modulus

Originally Posted by emperorzelos
No, it is a natural consequence of how the equation works.

If you have then out of neccesity . This is a trivial thing to prove so you cannot ever set c=a+b for it to work in fermats last theorem. The moment you set the equality, you are no longer dealing with fermals last theorem as we know that the equality must be false or it to relate to FLT.
I do prove the equality false in Fermat's last theorem via the Binomial Expansion for n >2.
(also for n=2, if one excludes Pythagorean triples; but not necessary for Fermat's proof.)
------------------------
iff

The point is that a and b must be independent (a,b); i.e., not on the same number line, but on orthogonal number lines. If they are on the same number line, then c =(a+b) = d, so ; that is, the identity of a and b is not preserved in the Binomial expansion because the lhs and rhs of c=(a +b) refer to the same single integer.

By the Binomial Expansion, true for all positive numbers, not just integers:

That is, (once again)

since rem(a,b,n) > 0

, n > 2 (Fermat's Theorem, my proof)

QED

(Try really, really hard to understand Fermat's Theorem AND the Binomial Expansion)

rem(a,b,n) is simply everything that is not or in the Binomial Expansion.
------------------
c < (a+b) merely says so

which is a true expression, but is a metric expression, not a proof of Fermat's Theorem, so I don't have to prove it, trivial as it is. It is irrelevant to the proof. (In particular, it is not true for all integers a, b, c, in particular for c > (a + b) - c cannot satisfy both equations.

c=(a+b) sets the foundation for the proof (by contradiction)

My proof is true for all integers (and numbers) in Fermat's equation. And I don't really need all the analysis in terms of vectors, the Binomial Theorem is just fine as it is. (I had been trying to justify the Binomial Theorem in terms of vectors by including Pythagorean triples for the case n = 2, but Fermat specifies n>2, so that analysis is also irrelevant to the actual proof).

6. ## Re: Proof of Fermat's Theorem by Modulus

Originally Posted by BuleriaChk
Even to the point of blatantly lying about my results in his own thread which he has now closed.

Grapes is also a fraud as a moderator.
Quote the blatant lies.

7. ## Re: Proof of Fermat's Theorem by Modulus

Originally Posted by BuleriaChk
I do prove the equality false in Fermat's last theorem via the Binomial Expansion for n >2.
(also for n=2, if one excludes Pythagorean triples; but not necessary for Fermat's proof.)
------------------------
iff

The point is that a and b must be independent (a,b); i.e., not on the same number line, but on orthogonal number lines. If they are on the same number line, then c =(a+b) = d, so ; that is, the identity of a and b is not preserved in the Binomial expansion because the lhs and rhs of c=(a +b) refer to the same single integer.

By the Binomial Expansion, true for all positive numbers, not just integers:

That is, (once again)

since rem(a,b,n) > 0

, n > 2 (Fermat's Theorem, my proof)

QED

(Try really, really hard to understand Fermat's Theorem AND the Binomial Expansion)

rem(a,b,n) is simply everything that is not or in the Binomial Expansion.
------------------
c < (a+b) merely says so

which is a true expression, but is a metric expression, not a proof of Fermat's Theorem, so I don't have to prove it, trivial as it is. It is irrelevant to the proof. (In particular, it is not true for all integers a, b, c, in particular for c > (a + b) - c cannot satisfy both equations.

c=(a+b) sets the foundation for the proof (by contradiction)

My proof is true for all integers (and numbers) in Fermat's equation. And I don't really need all the analysis in terms of vectors, the Binomial Theorem is just fine as it is. (I had been trying to justify the Binomial Theorem in terms of vectors by including Pythagorean triples for the case n = 2, but Fermat specifies n>2, so that analysis is also irrelevant to the actual proof).
Again, you cannot use binomial theorem assumes c=a+b which it cannot be in fermats last theorem.

is false, for example , then c=a+b or c=-(a+b) are valid.

I understand both binomial expansion and fermats last theorem and your thing doesn't logically follow. If you tried this in a mathematical journal they would laugh at you. You'd be sent to the crank archives.

UNderstand this, Malaria, Binomil expansions demand that c=a+b, but fermats last theorem demands that c<a+b

ergo they are not compatible as you do it as for one to tbe true, the other must be false and you cannot get them together.

8. ## Re: Proof of Fermat's Theorem by Modulus

Originally Posted by BuleriaChk
I do prove the equality false in Fermat's last theorem via the Binomial Expansion for n >2.
(also for n=2, if one excludes Pythagorean triples; but not necessary for Fermat's proof.)
------------------------
iff

The point is that a and b must be independent (a,b); i.e., not on the same number line, but on orthogonal number lines. If they are on the same number line, then c =(a+b) = d, so ; that is, the identity of a and b is not preserved in the Binomial expansion because the lhs and rhs of c=(a +b) refer to the same single integer.

By the Binomial Expansion, true for all positive numbers, not just integers:

That is, (once again)

since rem(a,b,n) > 0

, n > 2 (Fermat's Theorem, my proof)

QED
So, this is "true for all positive numbers, not just integers:"

, n > 2 (Fermat's Theorem, my proof)

It appears that your proof does show that.
(Try really, really hard to understand Fermat's Theorem AND the Binomial Expansion)

rem(a,b,n) is simply everything that is not or in the Binomial Expansion.
------------------
c < (a+b) merely says so

which is a true expression, but is a metric expression, not a proof of Fermat's Theorem, so I don't have to prove it, trivial as it is. It is irrelevant to the proof. (In particular, it is not true for all integers a, b, c, in particular for c > (a + b) - c cannot satisfy both equations.

c=(a+b) sets the foundation for the proof (by contradiction)

My proof is true for all integers (and numbers) in Fermat's equation. And I don't really need all the analysis in terms of vectors, the Binomial Theorem is just fine as it is. (I had been trying to justify the Binomial Theorem in terms of vectors by including Pythagorean triples for the case n = 2, but Fermat specifies n>2, so that analysis is also irrelevant to the actual proof).

9. ## Re: Proof of Fermat's Theorem by Modulus

Originally Posted by grapes
So, this is "true for all positive numbers, not just integers:"

, n > 2 (Fermat's Theorem, my proof)

It appears that your proof does show that.
You said your proof shows that. You couldn't be more wrong about it.

10. ## Re: Proof of Fermat's Theorem by Modulus

Originally Posted by grapes
You said your proof shows that. You couldn't be more wrong about it.
Time will tell. My .pdf is on my website.... and you're on record for everyone to see...

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