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

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 Joined: Thu Jul 24, 2014 9:00 am
Re: Analytical simple proof of Fermat's last theorem ????
I have not finished reading, but you have an error in your theorem.
1^{1}+3^{1}=4^{1}
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.
1^{1}+3^{1}=4^{1}
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.
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.

 Posts: 10
 Joined: Thu Jul 24, 2014 9:00 am
Re: Analytical simple proof of Fermat's last theorem ????
[quote="aruff"]I have not finished reading, but you have an error in your theorem.
1^{1}+3^{1}=4^{1}
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
1^{1}+3^{1}=4^{1}
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
Re: Analytical simple proof of Fermat's last theorem ????
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 wrote:well n=1 is a trivial case..... true for any additionaruff wrote:I have not finished reading, but you have an error in your theorem.
1^{1}+3^{1}=4^{1}
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
for n>=2 theorem is undoubtly true. I will change main theorem statement precisely for n>=2

 Posts: 10
 Joined: Thu Jul 24, 2014 9:00 am
Re: Analytical simple proof of Fermat's last theorem ????
Proof is very clear in the manuscript. Please read manuscript carefully.thundre wrote: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 wrote:well n=1 is a trivial case..... true for any additionaruff wrote:I have not finished reading, but you have an error in your theorem.
1^{1}+3^{1}=4^{1}
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
for n>=2 theorem is undoubtly true. I will change main theorem statement precisely for n>=2

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 Joined: Sun Jul 06, 2014 11:31 pm
Re: Analytical simple proof of Fermat's last theorem ????
The reasoning after (3.14) is not sufficient to prove that y_{1} and y_{2} are coprime. It could be the case that y_{1} = k g_{2} x_{2} and y_{2} = k g_{1} x_{1}, where k > 1 is coprime to both g_{1} and g_{2}. Without showing that y_{1} = g_{2}x_{2}, you cannot prove that rad(m) = rad(d_{1}d_{2}), and most of the rest of the proof fails.

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 Joined: Thu Jul 24, 2014 9:00 am
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
we have g1x1y1=g2x2y2
where gcd(g1x1,g2x2)=gcd(g1x1,y1)=gcd(g2x2,y2)=1
Re: Analytical simple proof of Fermat's last theorem ????
From those facts you have concluded that gcd(y1,y2)=1, but that does not follow from those facts alone.RanberSingh wrote:From where k come??? strange
we have g1x1y1=g2x2y2
where gcd(g1x1,g2x2)=gcd(g1x1,y1)=gcd(g2x2,y2)=1
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}

 Posts: 10
 Joined: Thu Jul 24, 2014 9:00 am
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.
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.

 Posts: 10
 Joined: Thu Jul 24, 2014 9:00 am
Re: Analytical simple proof of Fermat's last theorem ????
I once again thank you "jaap" for your important observation. I had wrongly concluded rad(m)=rad(d1d2) from m=ad2=bd1. But this somehow seems correct.
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.
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.

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 Joined: Sun Jul 06, 2014 11:31 pm
Re: Analytical simple proof of Fermat's last theorem ????
There is still a problem here. In Case 2, you say that d_{1}+d_{2}+2m (which is the same as c+m) cannot be an integer factor of d_{1}+d_{2}+m (which is the same as c). This is quite true (because d_{1}, d_{2}, and m are all positive), but it does not prove your point, because the left hand side is not c, but c^{n}. 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 wellknown results about Pythagorean triples, you can show that it actually is true that rad(m) = rad(d_{1}d_{2}). 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.
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 wellknown results about Pythagorean triples, you can show that it actually is true that rad(m) = rad(d_{1}d_{2}). 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.

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 Joined: Thu Jul 24, 2014 9:00 am
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).
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).

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 Joined: Sun Jul 06, 2014 11:31 pm
Re: Analytical simple proof of Fermat's last theorem ????
Say d_{1}=1, d_{2}=2, and m=6. Then f=3.
What if c=21? Then c+m=27. But then c+m is a factor of c^{3}.
So your proof still has a hole in it.
What if c=21? Then c+m=27. But then c+m is a factor of c^{3}.
So your proof still has a hole in it.

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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..
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.

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Re: Analytical simple proof of Fermat's last theorem ????
keep your such advice upto you
 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 FermatWiles’ theorem.
Re: Analytical simple proof of Fermat's last theorem ????
Please keep it friendly, otherwise I have to lock the topic.RanberSingh wrote:keep your such advice upto you
hk, moderator.

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Re: Analytical simple proof of Fermat's last theorem ????
I think I have acceptable reasoning to believe that rad(m)=rad(d1d2).
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).
Thus for n=2, odd n and n=2^k the proof is quite clear.
Only the reasoning for even n=(2^k)l, l an odd integer, is probably not good.
I will update the manuscript for any further comments.
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).
Thus for n=2, odd n and n=2^k the proof is quite clear.
Only the reasoning for even n=(2^k)l, l an odd integer, is probably not good.
I will update the manuscript for any further comments.