# Thread: Fermat's last, and mine too.

1. ## Re: Fermat's last, and mine too.

Originally Posted by BuleriaChk
Doesn't invalidate the proof at all. The proof specifies n>2.

The dot product is only valid for for the Pythagorean triple.
The dot product is *always* zero, no matter which positives integers a and b are, whether they are a part of a Pythagorean triple or not. You just can't do what you think you're doing. Your "proof" is invalid.
It is NOT valid for terms in rem(a,b,n) for n>2 Consider the case n=3, which has terms like and

In the terms containing products of a and b ( ), and , are NEVER orthogonal, and rem(a,b,3) only vanishes if a = 0 or b= 0.

(the terms only intersect at (0,0).

Village idiot...

2. ## Re: Fermat's last, and mine too.

Originally Posted by grapes
The dot product is *always* zero, no matter which positives integers a and b are, whether they are a part of a Pythagorean triple or not. You just can't do what you think you're doing. Your "proof" is invalid.
The dot product applies to Pythagorean triples for the case n=2. NOT for n>2 in the context of positive integers. If the vectors are not orthogonal in two dimensions the plane (a,b) or (x,y), then the dot product is not zero, since

It actually applies to all real numbers in a two dimensional Cartesian coordinate system, of which Pythagorean triples are a sublset.

as in the case n=2 only if

You're not only a Village Idiot, you're a Continental Idiot.

Try to get a refund from the school that taught you advanced math. Or even beginning math.

Hint: still reside on the (x,y) axes...

(Anyone not in the Peanut Gallery must be laughing their asses off).

3. ## Re: Fermat's last, and mine too.

Originally Posted by BuleriaChk
Village idiot...

_______________
Proof of Fermat's Last Theorem Updates 03/17/2017 9:28 PM PST
The pdf at that link in your signature still contains your erroneous "proofs". On the first page it says:
03./05/2017 07:05 AM PST Case n=2 dot product

It is important to understand that the Pythagorean Theorem means that the two legs of the right triangle and are orthogonal, so that For the Binomial Expansion, this means that
No, for positive numbers, is never equal to zero in the Binomial Expansion, is *never* equal to your

4. ## Re: Fermat's last, and mine too.

Originally Posted by grapes
The pdf at that link in your signature still contains your erroneous "proofs". On the first page it says:

No, for positive numbers, is never equal to zero in the Binomial Expansion, is *never* equal to your
The context is clear (well, ok, probably not to you); I am comparing the Pythagorean Triangle to the Binomial Expansion, not equating them.

Doesn't invalidate my proof, in any case (n>2)

(If Dr. Wiles contacts me about it, I might go back and clarify it...

5. ## Re: Fermat's last, and mine too.

Originally Posted by BuleriaChk
The context is clear (well, ok, probably not to you); I am comparing the Pythagorean Triangle to the Binomial Expansion, not equating them.
Doesn't invalidate my proof, in any case (n>2)
Of course it invalidates your "proof". For *any* solution to , n>1, c must be less than a+b. You never consider c<a+b, which is why your "proof" is so short--you've ignored the difficult part!
(If Dr. Wiles contacts me about it, I might go back and clarify it...
If you leave bonehead errors on the first page (or any page) you'll never get that chance.

6. ## Re: Fermat's last, and mine too.

Originally Posted by grapes

Of course it invalidates your "proof". For *any* solution to , n>1, c must be less than a+b. You never consider c<a+b, which is why your "proof" is so short--you've ignored the difficult part!

If you leave bonehead errors on the first page (or any page) you'll never get that chance.
Nah, I've presented the proof in so many other places in my pdf and on this forum, that the context is clear, and I don't consider any elements a,b,c that are not in vector format (which is implicitly understood in the context of my proof).

In the pdf quote you left out the context of the full sentence "... (orthogonal symbol) in the pair (a,b)" which automatically refers to the expression as a vector expression in Cartesian space You deliberately misquoted me out of context. Shame on you. You did not quote the full sentence. Despicable.

If you can't understand that the vector format is implicitly understood, then you're going to keep on being a village idiot, and your complaints are completely irrelevant to my proof. I am writing a lot of material that I am not posting directly to this forum, and I have no time for idiots that can't understand that in the context of the proof I am always talking about vectors. From the very start.

The Peanut Gallery is the only echo chamber that does not seem to understand this. Fricken' Village Idiots.

7. ## Re: Fermat's last, and mine too.

Originally Posted by BuleriaChk
Nah, I've presented the proof in so many other places in my pdf and on this forum, that the context is clear, and I don't consider any elements a,b,c that are not in vector format (which is implicitly understood in the context of my proof).

In the pdf quote you left out the context of the full sentence "... (orthogonal symbol) in the pair (a,b)" which automatically refers to the expression as a vector expression in Cartesian space You deliberately misquoted me out of context. Shame on you. You did not quote the full sentence. Despicable.
No, it was there:
Originally Posted by grapes
03./05/2017 07:05 AM PST Case n=2 dot product

It is important to understand that the Pythagorean Theorem means that the two legs of the right triangle a and b are orthogonal, so that For the Binomial Expansion, this means that
Again, I was just pointing out places where you claimed two expressions were equal, but it's easily shown that they're not equal. I don't care if you think you're justified in messing up the notation. If you ever cleaned up the notation, it'd be even more obvious that your "proof" is riddled with errors and misconceptions.

8. ## Re: Fermat's last, and mine too.

Originally Posted by grapes
No, it was there:

Again, I was just pointing out places where you claimed two expressions were equal, but it's easily shown that they're not equal. I don't care if you think you're justified in messing up the notation. If you ever cleaned up the notation, it'd be even more obvious that your "proof" is riddled with errors and misconceptions.
(minor edit of mine to clarify "comparision" which is totally unnecessary: Grapes omitted " in the pair (a,b)." which emphasizes the vector nature of the expression so my minor edit was unnecessary.

"It is important to understand that the Pythagorean Theorem means that the two legs of the right triangle and are orthogonal, so that For the Binomial Expansion, this means that comparing the Pythagorean triangle with the Binomial Expansion , in the pair (a,b)."

Fricken despicable Village Idiot, braying "hee haw" from the echo chamber of the Peanut Gallery.

9. ## Re: Fermat's last, and mine too.

Added a section for Astrophysicists to the FLT proof 03/19/2017 3:09 PM PST

10. ## Re: Fermat's last, and mine too.

Originally Posted by BuleriaChk
(minor edit of mine to clarify "comparision" which is totally unnecessary: Grapes omitted " in the pair (a,b)." which emphasizes the vector nature of the expression so my minor edit was unnecessary.

"It is important to understand that the Pythagorean Theorem means that the two legs of the right triangle and are orthogonal, so that For the Binomial Expansion, this means that comparing the Pythagorean triangle with the Binomial Expansion , in the pair (a,b)."

Fricken despicable Village Idiot, braying "hee haw" from the echo chamber of the Peanut Gallery.
So, your pdf repeats the same thing in the space of three lines, I leave off one of the repetitions and you complain I took the whole thing out of context because of it? C'mon

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