proof of sir fermats last theorem.?

Question

I have already proved in my 4th no. question that any number having any power can be expressed as difference of infinite sets of two perfect squares.now it is sir fermats last theorem {read a^(n)+b^(n)=c^(n) as a^(n)=c^(n)-b^(n) }.
we know , a^(n)=c^(2)-b^(2)
let c=x+e and b=x+d
here x is the common term between a,b and c and e and d are arbitrary constants according to the value of e and d the values of c,b and x will vary (values of e and d will be chosen by us)
putting the values of c and b we get
a^(n)=(x+e+x+d)(e-d) i.e.
a^(n)=(c+b)(e-d) _______________ (1)
here if we select e and d as c and b then we get
a^(n)=c^(2)-b^(2)
we know that in the formula of c^(n)-b^(n) is (c-b)(something)
but in the equation (1) the (c+b) term is there and if we select (e-d) as (c-b)(something) then we are not getting the formula of c^(n)-b^(n) but a^(n)=c^(2)-b^(2) can be written which is getting proved here also.So a^(n)=c^(n)-b^(n)is not is not having integer solutions for a,b and c for n >2

any queries are welcome but pls understand the thing carefully before asking.

"any number" means integers as well as the terminating decimals.

in the identity a^(n)=(c+b)(e-d) e and d can be varied and can be taken anything including
e=c and d=b but c and b cannot be varied (why?).

3 ^ (5) =122 ^ (2) - 121 ^ (2)
any given number with any given power can be expressed as difference of infinite sets of two perfect squares and those squares into difference of other two squares and so on.

Answer 1

your equation (1) a^(n) = (c+b)(e-d) is fine but e-d cannot be an integer since a^(n)/(c+b) is not an integer whereas a^(n)/(c-b) is an integer. In (1) if you select e and d to be c and b then you have a^(n) = c^2-b^2 but this is not adressing the original fermat equation. What you need to do is rewrite your argument in a more complete step-by-step form so it is perfectly clear what you are saying with respect to using the difference of squares. I realize you can't do that here with the limited sapce allowed. As it is, it is not clear or rigorous.

You have a creative flair but if a^(n)= c^(n)-b^(n) where n is odd (in fact n is an odd prime) then a^(n) = (c^(n/2)-b^(n/2))(c^(n/2)+b^(n/2)). Although a^n or any odd number can be expressed as the difference between 2 squares: (2^2)-(1^2) = 3; (3^2)-(2^2) =5; (4^2)-(3^2)= 7 etc. the values of any intergers q,r such that a^(n) = q^2-r^2 cannot be substituted for c and b. In fact if a^(n)= c^(n)-b^(n) then c-b divides a^n. In fact a^n = (c-b)(c^(n-1)-bc^(n-2)+b^2c^(n-3)-........). From your c=x+e and b=x+d it follows (c-b) =e-d = what I've written in brackets and we don't have a difference of squares.
A better elementary approach may be to consider the symmetric equation a^(n)+b^(n) = c^(n)+d^(n). For example 12^3+1^3 = 9^3+10^3 = 1729 and use the mutual divisibility properties of both sides. For larger n perhaps it may be shown that differences between the two sides may have a lower bound as a function of n, which would imply that 3 terms can only get so close. Just a thought.
But I encourage you to keep working on things you are interested in because you learn, and who knows you may be able to prove some things that have not been proven.

Answer 2

All of this makes no sense at all, your 4th. number question included. Sorry to be so direct

P.S. The sentence "x is the common term between a,b and c" is not an usually defined terminology, if you want to use it you have to define that, is it x=GCD(a,b,c) the Greatest Common Divisor? and so on... take the time to see what a theorem or any mathematical statement really means.

When you say, for example:
"we know , a^(n)=c^(2)-b^(2)"
you need to define which a,b,c are, are they any three integers? or you mean "let a,b,c are three integers verifing that equation"; otherwise you make only confusion for yourself and for others, That is not useful to you, certainly that is not mathematics. If you really love mathematics (as I can hint from your passion) take time to learn the commonly used mathematical language first, and the logic behind each math procedures.

I think you can do good things with a little bit more of humility and naturally with your passion linked to wisdom and clarity.
Thank you for your attention

Answer 3

It is well known that for about last 100 years, math departments of most universities usually ordered large supplies of a special form from typographies. It was printed in thousands of copies. This form usually looked like this:

"Dear Mr/Mrs/Ms ______!
We have received and analyzed your proof of Fermat's Last Theorem. It was a pleasure to read, and it is good to know that mathematics is becoming popular.
Unfortunately, you have an error in your proof on page ___, line ___, in the following statement: ____________. This statement is wrong, because of ___________ "

If not for such forms, the mathematicans would have to spend most of their time just sending replies to these proofs.

Answer 4

< a^(n)=c^(2)-b^(2)>>

If you make e and d no longer arbitrary, you haven't proven anything.

I just rethought that. You've actually proven that a = -2x, or a = 0 and b = c and d = e. 0 = 0.