Analytical simple proof of Fermat's last theorem ????

Primes, divisors, arithmetic, number properties, ...
RanberSingh
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Analytical simple proof of Fermat's last theorem ????

aruff
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Re: Analytical simple proof of Fermat's last theorem ????

I have not finished reading, but you have an error in your theorem.
11+31=41
Obviously 1 and 3 are both odd which means your main theorem is incorrect. Likely there is some reason for n>1. If there is no obvious reason for this, then there is a fault deeper in your proof.

ggoyo
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Re: Analytical simple proof of Fermat's last theorem ????

You state in the beginning of your paper that this theorem ($a$ and $b$ can't be both odd) will be proven, pretend that it leads to Fermat's theorem and then start something completely different in your paper.
Last edited by ggoyo on Fri Aug 08, 2014 10:04 am, edited 3 times in total. RanberSingh
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Re: Analytical simple proof of Fermat's last theorem ????

[quote="aruff"]I have not finished reading, but you have an error in your theorem.
11+31=41
Obviously 1 and 3 are both odd which means your main theorem is incorrect. Likely there is some reason for n>1. If there is no obvious reason for this, then there is a fault deeper in your proof

well n=1 is a trivial case..... true for any addition
for n>=2 theorem is undoubtly true. I will change main theorem statement precisely for n>=2

thundre
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Re: Analytical simple proof of Fermat's last theorem ????

RanberSingh wrote:
aruff wrote:I have not finished reading, but you have an error in your theorem.
11+31=41
Obviously 1 and 3 are both odd which means your main theorem is incorrect. Likely there is some reason for n>1. If there is no obvious reason for this, then there is a fault deeper in your proof
well n=1 is a trivial case..... true for any addition
for n>=2 theorem is undoubtly true. I will change main theorem statement precisely for n>=2
That changes the statement from false to unproven. If your proof relies on this statement, you must either provide a reference to an accepted proof of it or prove it yourself as a lemma. RanberSingh
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Re: Analytical simple proof of Fermat's last theorem ????

thundre wrote:
RanberSingh wrote:
aruff wrote:I have not finished reading, but you have an error in your theorem.
11+31=41
Obviously 1 and 3 are both odd which means your main theorem is incorrect. Likely there is some reason for n>1. If there is no obvious reason for this, then there is a fault deeper in your proof
well n=1 is a trivial case..... true for any addition
for n>=2 theorem is undoubtly true. I will change main theorem statement precisely for n>=2
That changes the statement from false to unproven. If your proof relies on this statement, you must either provide a reference to an accepted proof of it or prove it yourself as a lemma.

quantheory
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Re: Analytical simple proof of Fermat's last theorem ????

The reasoning after (3.14) is not sufficient to prove that y1 and y2 are coprime. It could be the case that y1 = k g2 x2 and y2 = k g1 x1, where k > 1 is coprime to both g1 and g2. Without showing that y1 = g2x2, you cannot prove that rad(m) = rad(d1d2), and most of the rest of the proof fails.

RanberSingh
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Re: Analytical simple proof of Fermat's last theorem ????

From where k come??? strange
we have g1x1y1=g2x2y2
where gcd(g1x1,g2x2)=gcd(g1x1,y1)=gcd(g2x2,y2)=1

jaap
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Re: Analytical simple proof of Fermat's last theorem ????

RanberSingh wrote:From where k come??? strange
we have g1x1y1=g2x2y2
where gcd(g1x1,g2x2)=gcd(g1x1,y1)=gcd(g2x2,y2)=1
From those facts you have concluded that gcd(y1,y2)=1, but that does not follow from those facts alone.
For example:
y1 = 2*3
g1x1 = 5*7
g2x2 = 3
y2 = 2*5*7
This satisfies gcd(g1x1,g2x2)=gcd(g1x1,y1)=gcd(g2x2,y2)=1 and g1x1y1=g2x2y2 but gcd(y1,y2)=2 (which is the k that quantheory was talking about).
Jaap's Puzzle Page RanberSingh
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Re: Analytical simple proof of Fermat's last theorem ????

thanks, I got it now what do you mean, but still main results will not change.
g1x1y1=g2x2y2, where gcd(g1x1,g2x2)=gcd(g1x1,y1)=gcd(g2x2,y2)=1
if gcd(y1,y2)=k then we can write
g1x1(kz1)=g2x2(kz2), where y1=kz1, y2=kz2 and gcd(z1,z2)=1,
still the result rad(m)=rad(d1d2) is true. Anyway I will check carefully again.

RanberSingh
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Re: Analytical simple proof of Fermat's last theorem ????

Anyway the main results are not strictly dependent on this relation, rad(m)=rad(d1d2).

It has not changed the main results of Fermat's Last Theroem in my article. I have to change partialy the main theorem. I still believe the main results are true. I hope you continue to read after the equation g1x1z1=g2x2z2. I uploaded a new version with some corrections.

quantheory
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Re: Analytical simple proof of Fermat's last theorem ????

There is still a problem here. In Case 2, you say that d1+d2+2m (which is the same as c+m) cannot be an integer factor of d1+d2+m (which is the same as c). This is quite true (because d1, d2, and m are all positive), but it does not prove your point, because the left hand side is not c, but cn. What you really need to show to complete your proof is that c+m contains at least one prime factor that is not also in c, i.e. that c ≠ 0 mod (rad(c+m)).

Since your proof no longer shows that m and c are coprime (because some of the factors in f may also be in c), you can no longer prove that c and c+m are coprime, and your proof is incomplete.

By the way, if you assume Fermat's Last Theorem to be true, combined with some well-known results about Pythagorean triples, you can show that it actually is true that rad(m) = rad(d1d2). But of course you can't use Fermat's Last Theorem to prove itself, so unless you can fix your proof of this statement, you can't use it...

Speaking for myself, it seems virtually impossible that such a straightforward proof of the theorem could emerge after all this time. I think that it's possible that a proof somewhat simpler than that of Wiles will be found one day, but I can't believe that anything this simple was missed by thousands of students and mathematicians for centuries.

RanberSingh
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Re: Analytical simple proof of Fermat's last theorem ????

I was wondering the same, how such a simple proof is missed for centuries by great mathematician who worked on this theorem for their whole life......... Even Andrew wiles also spent his whole prime time to find very complex proof of this theorem.

However, still your point is explainable. Since we have
gcd(c,d1d2)=gcd(f,d1d2)=1 and rad(m)=rad(fd1d2), d1 and d2 cannot be simultaneously equal to 1. So, c+m must have atleast one prime which is not in c or c^n. We know rad(c)=rad(c^n).

quantheory
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Re: Analytical simple proof of Fermat's last theorem ????

Say d1=1, d2=2, and m=6. Then f=3.

What if c=21? Then c+m=27. But then c+m is a factor of c3.

So your proof still has a hole in it.

RanberSingh
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Re: Analytical simple proof of Fermat's last theorem ????

wao,,, great to know. You are very good in math. I always missed basic points. But doing math of munbers is great fun for me. I do it sometimes in my free time. By profession I am physicist in material science.
Few months back I come to about Fermat's last theorem while reading a news about Beal's conjecture. Then I started thinking is it realy so hard to prove Fermat's last theorem.
I hope someone will find a answer in the procedure I have adopted. I also ceratnly will continue... whenever i find time. Thanks..

nicolas.patrois
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Re: Analytical simple proof of Fermat's last theorem ????

My little finger tells me that whenever someone finds an error, you will believe that the rest of your proof is valid despite it’s full of math errors that you should not commit. RanberSingh
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Re: Analytical simple proof of Fermat's last theorem ????

nicolas.patrois
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Re: Analytical simple proof of Fermat's last theorem ????

Heh, sorry, but I do not pretend that I have an elementary proof of Fermat-Wiles’ theorem. hk
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Re: Analytical simple proof of Fermat's last theorem ????

Please keep it friendly, otherwise I have to lock the topic.

hk, moderator. RanberSingh
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Re: Analytical simple proof of Fermat's last theorem ????

For n=2, we have $m=\sqrt(2d_1d_2).$ It implies that for n >= 3, we have m < \sqrt(2d_1d_2) for (d_1+m)^n+(d_2+m)^n=(d_1+d_2+m)^n because (d_1+m)^n, (d_2+m)^n and d_1+d_2+m)^n are strictly increasing with increasing n for given d_1 and d_2. We also clearly see in the manuscript that m^n==0 (\mod{d_1d_2}). Thus, rad(m)=rad(d_1d_2) because m^n==0 (\mod{d_1d_2}) and m <=\sqrt(2d_1d_2).