Dear Richard,

A person named George pointed out typing errors in mathematical formulas. They do not make the proof wrong. It is, just because while I write in papers I do not make errors. But the similarity in typing made me get mistakes. I am sorry, and here is the corrected one.

1. There is another explanation of a simple proof of Fermat’s last theorem as follows:

X^p + Y^p ?= Z^p (X,Y,Z are integers, p: any prime >2) (1)

2. Let‘s divide (1) by (Z-X)^p, we shall get:

(X/(Z-X))^p +(Y/(Z-X))^p ?= (Z/(Z-X))^p (2)

3. That means we shall have:

X’^p + Y’^p ?= Z’^p and Z’ = X’+1 , with X’ =X/(Z-X), Y’ = Y/(Z-X), Z’ =Z/(Z-X) (3)

4. From (3), we shall have these equivalent forms (4) and (5):

Y’^p ?= pX’^(p-1) + …+pX’ +1 (4)

Y’^p ?= p(-Z’)^(p-1) + …+p(-Z’) +1 (5)

5. Similarly, let’s divide (1) by (Z-Y)^p, we shall get:

(X/(Z-Y))^p +(Y/(Z-Y))^p ?= (Z/(Z-Y))^p (6)

That means we shall have these equivalent forms (7), (8) and (9):

X”^p + Y”^p ?= Z”^p and Z” = Y”+1 , with X” =X/(Z-Y), Y” =Y/(Z-Y), Z” =Z/(Z-Y) (7)

From (7), we shall have:

X”^p ?= pY’’^(p-1) + …+pY’’ +1 (8)

X”^p ?= p(-Z”)^(p-1) + …+p(-Z”) +1 (9)

Since p is a prime that is greater than 2, p is an odd number. Then, in (4), for any X’ we should have only one Y’ (that corresponds with X’) as a solution of (1), (3), (4), (5), if X’ could generate any solution of Fermat’s last theorem in (4).

By the equivalence between X’^p + Y’^p ?= Z’^p (3) and X”^p + Y”^p ?= Z”^p (7), we can deduce a result, that for any X” in (8), we should have only one Y” (that corresponds with X” ) as a solution of (1),(7),(8),(9), if X” could generate any solution of Fermat’s last theorem.

X” cannot generate any solution of Fermat’s last theorem, because we have illogical mathematical deductions, for examples, as follows:

i) In (8), if an X”1 could generate any solution of Fermat’s last theorem, there had to be at least two values Y”1 and Y”2 or at most (p-1) values Y”1, Y”2,…, Y”(p-1), that were solutions generated by X”, of Fermat’s last theorem. (Please note the even number (p-1) of pY”^(p-1) in (8)). But we already have a condition stated above, that for any X” we should have only one Y” (that corresponds with X”) as a solution of (1),(7),(8),(9), if X” could generate any solution Fermat’s last theorem. Fermat’s last theorem is simply proved!

ii) With X”^p + Y”^p ?= Z”^p, if an X”1 could generate any solution of Fermat’s last theorem, there had to be correspondingly one Y” and one Z” that were solutions generated by X”, of Fermat’s last theorem. But let’s look at (8) and (9), we must have Y” = -Z”. This is impossible by further logical reasoning such as, for example: We should have : X”^p + Y”^p ?= Z”^p , then X”^p ?= 2Z”^p or (X”/Z”)^p ?= 2. The equal sign, in (X”/Z”)^p ?= 2, is impossible. Fermat’s last theorem is simply again proved, with the connection to the concept of (X”/Z”)^p ?= 2. Is it interesting?

]]>A simple proof of Beal’s conjecture:

Now let’s start with A^x +B^y ?= C^z (1)

A,B,C are coprimes and x,y,z are natural numbers >2

Then we have: A^x + B^x + B^y –B^x –C^z ?=0 (2)

(A+B-B)^x +B^x + B^y-B^x –C^z ?=0 (3)

A^x + (B+A-A)^x +By-B^x – C^z ?=0 (4)

a) x is an odd number:

Then we get:

B^y-B^x –C^z + (A+B)^x -x(A+B)^(x-1).B+…+ x(A+B).B^(x-1) ?=0 (5)

B^y-B^x –C^z + (A+B)^x -x(A+B)^(x-1).A+…+ x(A+B).A^(x-1) ?=0 (6))

Let’s take any set of three coprime actual numbers A1, B1, C1 and partially replace A, B, C with A1, B1,C1 as follows:

B1^y-B1^x –C1^z + (A1+B1)^x -x(A1+B1)^(x-1).B+…+ x(A1+B1).B^(x-1) ?=0 (7)

B1^y-B1^x –C1^z + (A1+B1)^x -x(A1+B1)^(x-1).A+…+ x(A1+B1).A^(x-1) ?=0 (8)

Please note the common mathematical structure among (7) and (8). Both equations (7) and (8) are from an equation as follows:

B1^y-B1^x –C1^z + (A1+B1)^x -x(A1+B1)^(x-1).u+…+ x(A1+B1).u^(x-1) ?=0 (9)

Now we can easily see that the equation (9) has not any integer solution, because if there had been a solution U1 in (9), we had to get at least a set of three number U1 = A1 = B1 to replace U,A,B in (7),(8), (9) by U1, A1, B1. The replacement to get any solution with the condition U1 = A1 = B1 is impossible, with A1 and B1 are coprimes. Please note that if A=1, we then will get at least a triplet of U=A=B ?= 1, and the conjecture is also proved with 1+1 ?= C^z.

In other words, When A1 and B1 are coprime, either A1 > B1 or A1 < B1, then we can never get U1 = A1 = B1 to replace U, A,B in (7),(8),(9) to get any solution.

b) x is an even number: We can use a similar approach as the above approach, to prove the case x is even.

Fermat's last theorem can be proved by this method and some other simple proofs.

Pham Duc Sinh thaotrangtvt3@gmail.com

]]>A simple proof of Beal’s conjecture:

Now let’s start with A^x +B^y ?= C^z (1)

A,B,C are coprimes and x,y,z are natural numbers >2

Then we have: A^x + B^x + B^y –B^x –C^z ?=0 (2)

(A+B-B)^x +B^x + B^y-B^x –C^z ?=0 (3)

A^x + (B+A-A)^x +By-B^x – C^z ?=0 (4)

a) x is an odd number:

Then we get:

B^y-B^x –C^z + (A+B)^x -x(A+B)^(x-1).B+…+ x(A+B).B^(x-1) ?=0 (5)

B^y-B^x –C^z + (A+B)^x -x(A+B)^(x-1).A+…+ x(A+B).A^(x-1) ?=0 (6))

Let’s take any set of three coprime actual numbers A1, B1, C1 and partially replace A, B, C with A1, B1,C1 as follows:

B1^y-B1^x –C1^z + (A1+B1)^x -x(A1+B1)^(x-1).B+…+ x(A1+B1).B^(x-1) ?=0 (7)

B1^y-B1^x –C1^z + (A1+B1)^x -x(A1+B1)^(x-1).A+…+ x(A1+B1).A^(x-1) ?=0 (8)

Please note the common mathematical structure among (7) and (8). Both equations (7) and (8) are from an equation as follows:

B1^y-B1^x –C1^z + (A+B)^x -x(A+B)^(x-1).u+…+ x(A+B).u^(x-1) ?=0 (9)

Now we can easily see that the equation (9) has not any integer solution, because if there had been a solution U1 in (9), we had to get at least a set of three number U1 = A1 = B1 to replace U,A,B in (7),(8), (9) by U1, A1, B1. The replacement to get any solution with the condition U1 = A1 = B1 is impossible, with A1 and B1 are coprimes. Please note that if A=1, we then will get at least a triplet of U=A=B ?= 1, and the conjecture is also proved with 1+1 ?= C^z.

In other words, When A1 and B1 are coprime, either A1 > B1 or A1 < B1, then we can never get U1 = A1 = B1 to replace U, A,B in (7),(8),(9) to get any solution.

b) x is an even number: We can use a similar approach as the above approach, to prove the case x is even.

Fermat's last theorem can be proved by this method and some other simple proofs.

Pham Duc Sinh thaotrangtvt3@gmail.com

]]>Now let’s start with A^x +B^y ?= C^z (1)

A,B,C are coprimes and x,y,z are natural numbers >2

Then we have: A^x + B^x + B^y –B^x –C^z ?=0 (2)

(A+B-B)^x +B^x + B^y-B^x –C^z ?=0 (3)

A^x + (B+A-A)^x +By-B^x – C^z ?=0 (4)

a) x is an odd number:

Then we get:

B^y-B^x –C^z + (A+B)^x -x(A+B)^(x-1).B+…+ x(A+B).B^(x-1) ?=0 (5)

B^y-B^x –C^z + (A+B)^x -x(A+B)^(x-1).A+…+ x(A+B).A^(x-1) ?=0 (6))

Let’s take any set of three coprime actual numbers A1, B1, C1 and partially replace A, B, C with A1, B1,C1 as follows:

B1^y-B1^x –C1^z + (A1+B1)^x -x(A1+B1)^(x-1).B+…+ x(A1+B1).B^(x-1) ?=0 (7)

B1^y-B1^x –C1^z + (A+B)^x -x(A+B)^(x-1).A+…+ x(A+B).A^(x-1) ?=0 (8)

Please note the common mathematical structure among (7) and (8). Both equations (7) and (8) are from an equation as follows:

B1^y-B1^x –C1^z + (A+B)^x -x(A+B)^(x-1).u+…+ x(A+B).u^(x-1) ?=0 (9)

Now we can easily see that the equation (9) has not any integer solution, because if there had been a solution U1 in (9), we had to get at least a set of three number U1 = A1 = B1 to replace U,A,B in (7),(8), (9) by U1, A1, B1. The replacement to get any solution with the condition U1 = A1 = B1 is impossible, with A1 and B1 are coprimes. Please note that if A=1, we then will get at least a triplet of U=A=B ?= 1, and the conjecture is also proved with 1+1 ?= C^z.

In other words, When A1 and B1 are coprime, either A1 > B1 or A1 < B1, then we can never get U1 = A1 = B1 to replace U, A,B in (7),(8),(9) to get any solution.

b) x is an even number: We can use a similar approach as the above approach, to prove the case x is even.

Fermat's last theorem can be proved by this method and some other simple proofs.

Pham Duc Sinh thaotrangtvt3@gmail.com

]]>There is another explanation of a simple proof of Fermat’s last theorem as follows:

X^p + Y^p ?= Z^p (X,Y,Z are integers, p: any prime >2) (1)

Let‘s divide (1) by (Z-X)p, we shall get:

(X/Z-X)^p +( Y/Z-X)^p ?= (Z/Z-X)^p (2)

That means we shall have:

X’^p + Y’^p ?= Z’^p and Z’ = X’+1 , with X’ =(X/Z-X), Y’ =(Y/Z-X), Z’ =(Z/Z-X) (3)

From (3), we shall have these equivalent forms (4) and (5):

Y’^p ?= pX’^(p-1) + …+pX’ +1 (4)

Y’^p ?= p(-Z’)^(p-1) + …+p(-Z’) +1 (5)

Similarly, let’s divide (1) by (Z-Y), we shall get:

(X/Z-Y)^p +( Y/Z-Y)^p ?= (Z/Z-Y)^p (6)

That means we shall have these equivalent forms (7), (8) and (9):

X”^p + Y”^p ?= Z”^p and Z” = Y”+1 , with X” =(X/Z-Y), Y” =(Y/Z-Y), Z” =(Z/Z-Y) (7)

From (7), we shall have:

X”^p ?= pY”^(p-1) + …+pY” +1 (8)

X”^p ?= p(-Z”)^(p-1) + …+p(-Z”) +1 (9)

Since p is a prime that is greater than 2, p is an odd number. Then, in (4), for any X’ we should have only one Y’ (that corresponds with X’) as a solution of (1), (3), (4), (5), if X’ could generate any solution of Fermat’s last theorem in (4).

By the equivalence between X’^p + Y’^p ?= Z’^p (3) and X”^p + Y”^p ?= Z”^p (7), we can deduce a result, that for any X” in (8), we should have only one Y” (that corresponds with X’’ ) as a solution of (1),(7),(8),(9), if X” could generate any solution of Fermat’s last theorem.

X” cannot generate any solution of Fermat’s last theorem, because we have illogical mathematical deductions, for examples, as follows:

i) In (8), if an X”1 could generate any solution of Fermat’s last theorem, there had to be at least two values Y”1 and Y”2 or at most (p-1) values Y”1, Y”2,…, Y”(p-1), that were solutions generated by X”, of Fermat’s last theorem. (Please not the even number (p-1) of pY”^(p-1) in (8)). But we already have a condition stated above, that for any X” we should have only one Y” (that corresponds with X”) as a solution of (1),(7),(8),(9), if X” could generate any solution Fermat’s last theorem.

Fermat’s last theorem is simply proved!

ii) With X”^p + Y”^p ?= Z”^p , if an X”1 could generate any solution of Fermat’s last theorem, there had to be correspondingly one Y” and one Z” that were solutions generated by X”, of Fermat’s last theorem. But let’s look at (8) and (9), we must have Y” = -Z”. This is impossible by further logical reasoning such as, for example:

We should have : X”^p + (-Z”)^p ?= Z”^p, then X”^p ?= 2 Z”^p, or (X”/Z”)^p ?= 2. The equal sign, in (X”/Z”)^p ?= 2, is impossible.

Fermat’s last theorem is simply again proved, with the connection to the concept of (X”/Z”)^p ?= 2. Is it interesting? ]]>