From: bassam king karzeddin on
> Sure...
>
> Case 1 has been known for about 100 years before you
> were born (the website)-- so you might consider that
> you are in good company and that you are taking
> credit for what rightfully belongs to others...
>
Hello Joe

It seems that you know much better than (me & many)
Who are you sir

I really did it my self as an interest knowing that I'm a civil engineer and didn't study advanced courses in mathematics, but any way I'm so glad it was there befor I was born, also addeing any suitable references or web site will be very much appreciated.

> Case 2 was not so easy and Wiles had to go out of his
> way to "create" the math to solve it - so unless you
> can show how you had it in about 1985, you are not
> likely to get much credit.
>
> Joe

Do you mean by case 2, that when n is even positive integer, if so then it is too simple, because it is another picture of case 1 only.

but, if you mean by case 2, when there are coefficients (a & b) that may not be equal to 1.

or do you mean that when the

variable is > (m^m)*(k^k)/(n^n), which I think so, then

Please do explain or

make sure they belong to me with many evidances befor Wiles proofs, YES BEFOR WILES PROOFS

The strongest is my licenced book with copy right note, and really they don't require more than few pages to derive with a very ELEMENTARY methods

The other many proofs are the many replies that I do have (many copies with many friends that I will provide any one interested by post office box) from the very reputable JOURNALS WITH EXACT DATES, WITH SIGNUTURES AND NAMES I have already mentioned in this and other threads, but unfortunatly IT WAS NOT SUITABLE FOR ALL OF THEM, WITH OUT MENTIONING ANY REASON, and that only made me thinking I may be a mad man

Of course I didn't associate my formula to any other problem as FLT, because simply I never heard about it or about any unsolved problem befor the year 2004.

My Regards
B.Karzeddin
From: bassam king karzeddin on
> Yes to both. The best is to find the original
> announcements in 199x. This forum has some emails
> from Wiles and others who give shorter explanations.
> It took Wiles 200 pages to do what you are trying to
> do in one line.
>
> Joe

Hello Joe

I'm so glad you are convenced about (my or any one befor me) formula, reminding others it is a FORMULA AND NOT A CONJECTURE, however it took me so long to derive it.

I think also you got the complete proof in that ONE line only.

I wish also to see those e-mails from Wiles.

about your cubic equation 199x,
make sure it is a quadratic form because
(x=0, x=1, x=4) only

Many Regards
B.Karzeddin
From: Joseph Parranto on
Let me be very clear that I cannot comment on your formulas...I do not have enough skill there.
If you search out the proofs of others, and actually read them, you will be able to tell if they are copies of yours. But only you can do that - no one else.

Wiles Proof can be downloaded from the publisher, and there are many commentaries on it and explanations from many sources. You have access to the whole world from your computer - time to use it to search.

Case 1 was proved in 1910...and in 1930 the basis of Wiles proof was established but never finished. Your work is like that - unfinished = Not complete.

If you are worried about theft, you are definitely in the wrong business - intellectual property is almost forever. Publish everything you have completely for others to evalutate it. Your claims are not of any value to anybody unless they are available...

JFP
From: jiahao_anti-addictgamer on
On Feb 21, 10:51 pm, bassam king karzeddin <bas...(a)ahu.edu.jo> wrote:
> Fermat's Last theorem short proof
>
> We have the following general equation (using the general binomial theorem)
>
> (x+y+z)^p =p* (x+y)*(x+z)*(y+z)*N(x,y,z) + x^p+y^p+z^p
>
> Where
> N (x, y, z) is integer function in terms of (x, y, z)
> P is odd prime number
> (x, y, z) are three (none zero) co prime integers?
>
> Assuming a counter example (x, y, z) exists such that (x^p+y^p+z^p=0)
>
> (x+y+z)^p =p* (x+y)*(x+z)*(y+z)*N (x, y, z)
>
> CASE-1
> If (p=3) implies N (x, y, z) = 1, so we have
>
> (x+y+z)^3 =3* (x+y)*(x+z)*(y+z)
>
> Assuming (3) does not divide (x*y*z), then it does not divide (x+y)*(x+z)*(y+z),
> So the above equation does not have solution
> (That is by dividing both sides by 3, you get 9 times an integer equal to an integer which is not divisible by 3, which of course is impossible
> I think proof is completed for (p=3, and 3 is not a factor of (x*y*z)
>
> My question to the specialist, is my proof a new one, more over I will not feel strange if this was known few centuries back
>
> Thanking you a lot
>
> Bassam King Karzeddin
> Al-Hussein Bin Talal University
> JORDAN


the number of pages were big . BUT I would just like t ask this
question , does anyone of you have
a verified proof of FLT ? ( made by yourselves or others except Wiles)


Thank you,
John Howard Sim
14 / 3 / 2007

From: Gerry Myerson on
In article
<11558709.1173924022873.JavaMail.jakarta(a)nitrogen.mathforum.org>,
Joseph Parranto <jparranto(a)yahoo.com> wrote:

> Let me be very clear that I cannot comment on your formulas...I do not have
> enough skill there.
> If you search out the proofs of others, and actually read them, you will be
> able to tell if they are copies of yours. But only you can do that - no one
> else.
>
> Wiles Proof can be downloaded from the publisher, and there are many
> commentaries on it and explanations from many sources. You have access to the
> whole world from your computer - time to use it to search.
>
> Case 1 was proved in 1910...and in 1930 the basis of Wiles proof was
> established but never finished. Your work is like that - unfinished = Not
> complete.

Say, what?

If by Case 1 you mean what used to be called The First Case of Fermat's
Last Theorem, that is, the case where the prime exponent does not divide
any of the bases, then I think you're wrong about it having been proved
in 1910. What's your source?

As for your comment about 1930, I have no idea what you are talking
about.

--
Gerry Myerson (gerry(a)maths.mq.edi.ai) (i -> u for email)