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August 10th, 2014, 10:15 PM   #131
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An elegant proof is for sure more direct. Algebraic division can solve, but is still not clear to 99% of readers...

Once again: is just question to linearize and transfrom in 2 triangles:

One of area: A^2
The other of area: B^2 = C^2- A^2

If you correctly linearize the problem you'll see that the 2 triangle are not similar for the conditions A,B,C integers and coprimes, hence the proof.

But the most elegant point is that from n>=3 the derivate is a curve, for so no way to square in the integers... or (for n=3):

can be rewritten for sure for

sorry is trivial now...

BUT IF WE WRITE LIKE FERMAT is asking himself : is possible that what rest is again a (same) power of another integer ?

But to do that without make mistake we can just write:



This is for sure always true.

BUT if we wanna rest in the integers we can just correctly re-write (and ask):


I have to remember once again that this is a play between sum and integrals. now we well know that the one become the other passing to the limit. Probably this is also well clear to Fermat, that also very well know ALSO the properties of the numbers so that is also possible to have again a power using a sum like this: A^3 = 1+7+19+... 3x^2-3x+1 or the most general case a STEP SUM like this: A^3 = 1/k + 3x^2/k -3x/k^2+1/k^3 if we jump in x with step 1/k, so x1=1/k, x2= 2/k etc..

SO the (1) IS NO LONGER A CONTINUE FUNCTION SINCE YOU'VE A GAP IN CASE YOU LOOK TO THE AREA BELLOW THE DERIVATE (that viceversa is continue) FROM "A" AND "B+1". Since we knoe the properties of Sum and Integrals, this is enought to say; NO SOLUTION, but to prove we can still ask (and correctly write a sum step 1 in the integers, or a step sum that is more precise, as precise as we want (1/k), till the limit k->infinite)

So can be possible to find a solution in the integers for n>=3 ?

But this GEAR can't work, or has no solution if A,B,C are integers and coprimes since under a continue derivate curve y'= 3x^2 (in this case) we can continue to make littlest and littlest columns, (or triangles) but we can't find the solution since the gear teeth of C^n-B^n are not similar to the one of A^n and C^n (or the two area coverted in two triangleas, are not similar):

(I remember I use a mine special "Step sum", if you don't like pls trasnfrom in a normal sum with changeing the variable in the sum so have the sum from 1 to.... and not from 1/k step 1/k as I wrote here when you see lower limit 1/k)

So rising K in N we found no solution since A,B,C coprimes... so

If you've to understood what I mean for step sum we cannot complete (or square in "N" or "Q") the area bellow the curve till we go to the limit or we write:

so we have to accept 2 integers and one (A in this case) NON integer. This is the Dedekind section definition for an irrational.


here what we can do with a linear derivate:

and what we cannot (if we won't use the limit so we rest in N or in Q):

Last edited by complicatemodulus; August 10th, 2014 at 10:52 PM.
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October 8th, 2014, 12:00 PM   #132
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Once more,all numbers involved are natural
What we already know
$\displaystyle X^N+Y^N=Z^N$ (1), N┼XYZ, X=NA+a, Y=NB+b, Z=NC+c (a,b,c are the remainders of divisions X/N, Y/N and Z/N respectively).Of course N┼abc
a+b=c (2), $\displaystyle N^2|aY-bX$, $\displaystyle N^2|aZ-cX$, $\displaystyle N^2|bZ-cY$ (relationships (3))
Sentences used:
Sentence I:$\displaystyle N^m|D-E \rightarrow N^{m+1}|D^N-E^N$ and $\displaystyle N^m|D+E \rightarrow N^{m+1}|D^N+E^N$
Sentence II: $\displaystyle N|a_1a_2…a_m\rightarrow N|a_1$ or $\displaystyle N|a_2$ or …or $\displaystyle N|a_m$
Sentence III: If $\displaystyle N^kK=N^mM+N^pP$ with k>m, k>p and N┼MP, then m=p. The proof is rather simple, assuming e.g. m<p and dividing the relationship by $\displaystyle N^m$.

Corrolary I. If If $\displaystyle N^m|aY-bX$ and $\displaystyle N^m|aZ-cX$ then $\displaystyle N^m|X+Y-Z$. The proof is quite simple
Actually,if two of the following 4 conditions are true, so are the other two: 1st: $\displaystyle N^m|aY-bX$, 2nd: $\displaystyle N^m|aZ-cX$ 3rd: $\displaystyle N^m|bZ-cY$, 4th $\displaystyle N^m|X+Y-Z$.

Theorem I If $\displaystyle N^m|aY-bX$ and $\displaystyle N^m|aZ-cX$ then $\displaystyle N^{m+1}|a^N+b^N-c^N$
Proof: Sentence I : $\displaystyle N^m|aY-bX\rightarrow N^{m+1}|a^NY^N-b^NX^N$ and likewise: $\displaystyle N^{m+1}|a^NZ^N-c^NX^N$
Substracting and as $\displaystyle X^N+Y^N=Z^N$: $\displaystyle N^{m+1}|a^NX^N+b^NX^N -c^NX^N \rightarrow\ (sentence II) N^{m+1}|a^N+b^N -c^N$
Actually,if two of the following 4 conditions are true, so are the other two: 1st: $\displaystyle N^m|aY-bX$, 2nd: $\displaystyle N^m|aZ-cX$ 3rd: $\displaystyle N^m|bZ-cY$, 4th $\displaystyle N^{m+1}|a^N+b^N-c^N$.

Theorem II If $\displaystyle N^m|aY-bX$ and $\displaystyle N^m|aZ-cX$, then $\displaystyle N^{m+1}|a^{N-1}-b^{N-1}$, $\displaystyle N^{m+1}|a^{N-1}-c^{N-1}$, $\displaystyle N^{m+1}|c^{N-1}-b^{N-1}$
Proof: First a memo: by notating $\displaystyle D^E||F$ we mean $\displaystyle D^E|F$ but $\displaystyle D^{E+1}\dagger F$ .
So let $\displaystyle N^q||a^{N-1}-b^{N-1}=N^qQ$ , $\displaystyle N^r||c^{N-1}-b^{N-1}=N^rR$ and $\displaystyle N^u||a^{N-1}-c^{N-1}=N^uU$ (of course N┼QRU).We know q,r and u are natural numbers (Fermat’s little theorem).Assuming they are all smaller than m+1,we’ll end to absurdity.
By theorem I $\displaystyle N^{m+1}|a^N+b^N-c^N=aa^{N-1}+b^N-cc^{N-1} =
a(N^qQ+b^{N-1})+b^N-c(N^rR+b^{N-1}) = N^qQa-N^rRc+b^{N-1}(a+b-c) = N^qQa-N^rRc$ (4).
Since N┼Qa and N┼Rc, by sentence III:q=r .Likewise q=r=u and (4) turns to $\displaystyle N^{m+1}| N^q(aQ-cR)\rightarrow N^{m+1-q}| aQ-cR$
Likewise $\displaystyle N^{m+1-q}| cU-bR$ .
Substracting these last two and because a+b=c: $\displaystyle N^{m+1-q}| cU-aQ+aR$ (5)
But $\displaystyle a^{N-1}-b^{N-1}=c^{N-1}-b^{N-1}+a^{N-1}-c^{N-1}\leftrightarrow N^qQ=N^qR+N^qU\rightarrow Q=R+U$ which turns (5) to $\displaystyle N^{m+1-q}| cU-aU\rightarrow (sentence II) N^{m+1-q}| c-a=b$, absurd
Therefore q=r=u≥m+1. Hence: $\displaystyle N^{m+1}|a^{N-1}-b^{N-1}$, $\displaystyle N^{m+1}|a^{N-1}-c^{N-1}$ , $\displaystyle N^{m+1}|c^{N-1}-b^{N-1}$

Theorem III: If $\displaystyle N^m|aY-bX$ (6) and $\displaystyle N^m|aZ-cX$, then $\displaystyle N^m|X^{N-1}-Y^{N-1}$, $\displaystyle N^m|X^{N-1}-Z^{N-1}$ , $\displaystyle N^m|Z^{N-1}-Y^{N-1}$
Proof: $\displaystyle N^m|aY-bX\rightarrow (sentence\ I) N^{m+1}|a^NY^N-b^NX^N$ (7) and (theorem II) $\displaystyle N^{m+1}|a^{N-1}-b^{N-1}\rightarrow N^{m+1}|a^Nb-ab^N = N^{m+1}K$ and so by (7):
$\displaystyle N^{m+1}|(N^{m+1}K+ab^N)Y^N/b-b^NX^N\rightarrow N^{m+1}| ab^NY^N-b^{N+1}X^N\rightarrow (sentence\ II) N^{m+1}| aY^N-bX^N =$ (by (6)) $\displaystyle (N^mL+bX)Y^{N-1}-bX^N\rightarrow N^m| bXY^{N-1}-bX^N\rightarrow (sentence\ II) N^m| Y^{N-1}-X^{N-1}$ Likewise: $\displaystyle N^m|X^{N-1}-Z^{N-1}$ , $\displaystyle N^m|Z^{N-1}-Y^{N-1}$

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January 15th, 2015, 09:21 AM   #133
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While I’m banging my head on the wall to help ideas come about the first case of FLT, I thought we might say some words about the second case, when N|XYZ in $\displaystyle X^N+Y^N=Z^N$

In the text, symbol ┼ means “does not divide” and rel.means relationship
Sentences to use:
Sentence I: If N is a prime and N|ab, then N|a or N|b. It holds for more than two factors a and b
Sentence II: With E,D and F any natural numbers E┼D-F & $\displaystyle (E,D^6+F^6)=1$, then
$\displaystyle E ┼ D^{6Μ-5}-D^{6Μ-6}F+D^{6Μ-11}F^6-D^{6Μ-12}F^7+…+D^7F^{6Μ-12}-D^6F^{6Μ-12}+DF^{6Μ-6}-F^{6Μ-5}$
The proof lies in #125 post of this thread (theorem II there)

So,assume N|Z with no loss of generality.Of course N doesn’t divide X or Y, or X,Y,Z wouldn’t be pairwise prime
Then it is known that $\displaystyle N^2|X^{N-1}-1$, $\displaystyle N^2|Y^{N-1}-1$, $\displaystyle N^2|X+Y$ (1),
If N=6q+1 the first two of the rel.above lead to:
$\displaystyle N^2|(X^4+X^2Y^2+Y^4)(X^{N-6}-X^{N-7}Y+X^{N-12}Y^6-X^{Ν-13}Y^7+…+X^7Y^{Ν-13}-X^6Y^{Ν-12}+XY^{N-7}-Y^{N-6)}$ (2)
For details on how this long achived,see #125 post again
On the other hand,due to (1),it’s elementary to show:
$\displaystyle N|X^K+Y^K$ (3) for every odd K
If (3) held for some even K,we would have:
$\displaystyle N|X^K+Y^K =Nm$ and as K+1 is odd,
$\displaystyle N|X^{K+1}+Y^{K+1}=XX^K+Y^{K+1}= X(Nm-Y^K)+Y^{K+1} \rightarrow N| -XY^K+Y^{K+1} = Y^k(Y-X)$
and since N┼Y we imply $\displaystyle N\dagger Y^K$ ,therefore sentence I commands:
N| Y-X absurd, because then,due to (1) N|Y & N|X
So $\displaystyle N\dagger X^6+Y^6$ and as N is a prime $\displaystyle (N,X^6+Y^6)=1$
Given that and since N-6=6q-5, after sentence II we conclude:
$\displaystyle N┼ X^{N-6}-X^{N-7}Y+X^{N-12}Y^6-X^{Ν-13}Y^7+…+X^7Y^{Ν-13}-X^6Y^{Ν-12}+XY^{N-7}-Y^{N-6}$
Then (2) and sentence I result to:
$\displaystyle N^2| X^4+X^2Y^2+Y^4 = (X^2+XY+Y^2)(X^2-XY+Y^2) \rightarrow (sentence I) N| X^2+XY+Y^2$ or $\displaystyle N|X^2-XY+Y^2$
So if $\displaystyle N|X^2+XY+Y^2 = (X+Y)^2-XY$ and due to (1):N|XY → N|Y or N|X, absurd
On the other hand if $\displaystyle N|X^2-XY+Y^2 = (X+Y)^2-3XY$ and again N|3XY and as N≠3 sentence I leads us to the same absurdity
Conclusion: All 6q+1 primes satisfy 2nd case of FLT,as they do with the 1st one
Or I'm wrong.In which case I'll change my name to crankshaft
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January 20th, 2015, 12:49 AM   #134
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Sorry for borken link in the previous post (probably nobody care of it...):

Here one of the way to transform powers of integers in trapezoid, and what happen in FLT trick (I still believe in graph visualization to show immediately what we are doing, still if it's a very hard work...):

Sorry there is some erro on the lower picture: the height of the bigger trapezoid is of course C (not A that is the height of the littlest one ).
Also the eight of the new trapezoid Taonc must be expressed in term of C and Tan(Gamma) where Gamma is the angle....

Is easy to have the abscissa X, or the eight X of the new trapezoid Taonc, with a simple identity:

Exceding area over X = Missing Upper Area between 0 and X

This two areas are not proportional for FLT request, so no way to tessellate with a common non infimus triangle.

And this is true for any n>=3

All the 2th degree equation that solve X (just parameters will change) gives back an irrational, so Tb cannot be an integer area, so no way for B to be an integer.

Nothing change if we decide to strech all the trapezoides in the abscissa x multiply it by an integer K... (more concerning are necessary).

Last edited by complicatemodulus; January 20th, 2015 at 01:00 AM.
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March 7th, 2015, 10:14 PM   #135
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For a number of reasons this thread has run its course.We'll share some new thoughts we have in a very short while
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March 8th, 2015, 10:37 AM   #136
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Where is the error now:

I fond that to respect the rule of the sum that gives powers:

(example for n=3) Fermat ask if:


so he ask if:


But we know that :

and that for (my ;-P) rule of the sum we can lower both the limits of the second sum without changeing the result from A+1 to 1 with this trick:


So since the (3) is for sure right the (2) will be wrong or:

while Fermat ask if (applying on the (2) the same rule as in the (3)):

Cvd... since A<B.

It works for n=2 since the term of the sum is linear so it is possible to arrange the trick to be a square... Not if the complicate modulus has others non linear terms

This is what I mean when I say: derivate = curve imply Fermat is right... (but I for sure made some mistake....)
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