# Thread: Proof of Fermat's Theorem using vectors

1. ## Re: Proof of Fermat's Theorem using vectors

Originally Posted by BuleriaChk
{}^{ }As far as Grapes' contention that there is something wrong with c = a + b, I would point out that its extension to the Binomial expansion is agnostic as far as the ordeing is concerned.

That is c=(a+b), a=(c+b), b = (a + c) all work in the expansion, since rem(a,b,n), rem(c,b,n) and rem(a,c,n) all are greater than zero - that is, if you pick any three positive integers, it doesn't matter which ones go where in that equation. The ordering is agnostic.

So statements like "c <(a+b)" are just as irrelevant a<(c+b) or b <(a+c). That is why these statements can't be true for all integers. c<(a+b) does not include the set c>(a+b) so the inequalities are irrelevant.

My proof works because

IFF works just as well as
IFF and
IFF

for ANY three positive integers a,b, c no matter what the order in the expression.
And if that is true, the rest follows because rem(x,y,n) > 0 ALWAYS.

I don't have to express the proof in vectors; but the relativistic unit circle is very interesting to me.

The only issue is that a and b (or whiever elements appear in the sum) represent different integers in the expression without any other variables or constants.

(So, for example, in the first case a=b says c = 2a = d, so since they are on the same number line. That is why I would use the notation (a,b) or (c,b) or (a,c) respectively in the examples above, since they indicate different number lines (cartesian coordinates).

That said, I have been inconsistent technically in distinguishing between vector spaces and coordinate systems, but I think the distinction and usage are generally clear from the context. There are imporatnt difference between coordinate systems and vectors.

But if one doesn't know what a coordinate system is to begin with, the question is moot....
Your if and only if statements are false, set n=2m and the negative answers will be just as valid.

2. ## Re: Proof of Fermat's Theorem using vectors

Originally Posted by emperorzelos
Your if and only if statements are false, set n=2m and the negative answers will be just as valid.
He specified three positive integers.

3. ## Re: Proof of Fermat's Theorem using vectors

Originally Posted by grapes
He specified three positive integers.
If that is the case is it a trivial thing which makes it very unteresting, it is like saying well no shit sherlock.

4. ## Re: Proof of Fermat's Theorem using vectors

Originally Posted by emperorzelos
If that is the case is it a trivial thing which makes it very unteresting, it is like saying well no shit sherlock.
All math is trivial

I do use that, in the last step of my argument, post #9 of this thread.

5. ## (Another) Proof of Fermat's Theorem (which is actually the same)

The context of the analysis below can also be compared with the relativistic approach in which rem(a,b,2) corresponds to a Lorentz "boost" in addition to a Lorentz rotation for two integers.

The Relativistic Unit Circle

is the Binomial Expansion for the case n=2.

is the equation of a circle.

let r, x, and y be integers, a=x, b=y, c=r.

is the equation of all Pythagorean circles where the radius c is an integer and the sides a and b form a Pthagorean right triangle within the circle.

If the graph is not a circle, the Binomial expansion yields

for the case n=2, which is the equation of a Pythagorean circle and a rectangle of area = 2ab
---------------------
There are now 4 elements: on the lhs, with three integers: , and 2ab - on the r.h.s., so cannot be an integer used to construct Fermat's equation since all the Pythagorean triples require that 2ab = 0. This also follows from the vector analysis and the relativistic unit circle. So c cannot be part of a Pythagorean triple in the Binomial Expansion for the case n=2, and thus is not an integer.

The Binomial theorem is valid for all n > 2 where rem(a,b,n) > 0. Fermat's Theorem is proved; c is not an integer, so is not an integer.
---------------------
I can't believe the urban legend that no-one has proven Fermat's theorem isn't a fricken joke on all of us .....

6. ## Re: (Another) Proof of Fermat's Theorem (which is actually the same)

Originally Posted by BuleriaChk
I can't believe the urban legend that no-one has proven Fermat's theorem isn't a fricken joke on all of us .....
That is because mathematicians are much smarter than you and know binomial theorem is not applicable to fermats last theorem.

7. ## Re: Proof of Fermat's Theorem using vectors

I have added a connection to Fermat's theorem to my discussion of the relatavistic unit circle at:

Relativistic Unit Circle

which shows the connection to both the Pythagorean circle and the Binomial Theorem (via a Lorentz "boost")

It shows that the proof of Fermat's Theorem from STR and the Lorentz transform is consistent with that of the Binomial theorem.

In particular, the Lorentz "boost" at is consistent with the requirement of the Binomial Therorem that for a Pythagorean triple (or any triangle inscribed in the Relativistic unit circle, since and are independent (orthogonal) in the first (positive) quadrant.

The vector analysis from the OP is still being added to the pdf at present, but is not necessary, as the relation is clear geometrically.

8. ## Re: Proof of Fermat's Theorem using vectors

Originally Posted by BuleriaChk
I have added a connection to Fermat's theorem to my discussion of the relatavistic unit circle at:

Relativistic Unit Circle

which shows the connection to both the Pythagorean circle and the Binomial Theorem (via a Lorentz "boost")

It shows that the proof of Fermat's Theorem from STR and the Lorentz transform is consistent with that of the Binomial theorem.

In particular, the Lorentz "boost" at is consistent with the requirement of the Binomial Therorem that for a Pythagorean triple (or any triangle inscribed in the Relativistic unit circle, since and are independent (orthogonal) in the first (positive) quadrant.

The vector analysis from the OP is still being added to the pdf at present, but is not necessary, as the relation is clear geometrically.
Then explain how c=a+b and c<a+b can both be true at the same fucking time. How can a number be less than, but not equal, to another number, and at the same time be equal to the latter number?

And still you do not know fucking notation.

9. ## Re: (Another) Proof of Fermat's Theorem (which is actually the same)

Originally Posted by BuleriaChk
The context of the analysis below can also be compared with the relativistic approach in which rem(a,b,2) corresponds to a Lorentz "boost" in addition to a Lorentz rotation for two integers.

The Relativistic Unit Circle

is the Binomial Expansion for the case n=2.

is the equation of a circle.

let r, x, and y be integers, a=x, b=y, c=r.

is the equation of all Pythagorean circles where the radius c is an integer and the sides a and b form a Pthagorean right triangle within the circle.

If the graph is not a circle, the Binomial expansion yields

for the case n=2, which is the equation of a Pythagorean circle and a rectangle of area = 2ab
---------------------
There are now 4 elements: on the lhs, with three integers: , and 2ab - on the r.h.s., so cannot be an integer used to construct Fermat's equation since all the Pythagorean triples require that 2ab = 0.
Nonsense

By definition, a Pythagorean triple is three nonzero integers a,b,c such that , so 2ab is *never* zero.

This has been pointed out before, and you keep ignoring it.
This also follows from the vector analysis and the relativistic unit circle. So c cannot be part of a Pythagorean triple in the Binomial Expansion for the case n=2, and thus is not an integer.
Well, you've said c=a+b, so of course a,b,c is not a Pythagorean triple--they don't satisfy the Triangle Inequality. But, c is an integer, it's equal to a+b
The Binomial theorem is valid for all n > 2 where rem(a,b,n) > 0. Fermat's Theorem is proved; c is not an integer, so is not an integer.
Pure nonsense
---------------------
I can't believe the urban legend that no-one has proven Fermat's theorem isn't a fricken joke on all of us .....

10. ## Re: (Another) Proof of Fermat's Theorem (which is actually the same)

Originally Posted by grapes
Nonsense

By definition, a Pythagorean triple is three nonzero integers a,b,c such that , so 2ab is *never* zero.

This has been pointed out before, and you keep ignoring it.

Well, you've said c=a+b, so of course a,b,c is not a Pythagorean triple--they don't satisfy the Triangle Inequality. But, c is an integer, it's equal to a+b

Pure nonsense
Neither Grapes or the Naked Emperor have gotten to the point in their studies where manknind has invented the wheel.
Or even the square. They only think in one dimension, clapping sticks together to measure them....
(John Gabriel had that problem as well....)

Even the Greeks knew about the circle....

But the Relativistic circle, and the characterization of integer generation by Lorentz rotation is at the foundation of sampling theory, signal processing and analysis and analysis, (For the Lorentz Boost, imagine that the invariant initial condition ct is replaced by Planck's constant, and then ask yourself what happens to and h as goes to infinity.

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