# Fermat’s Last Theorem Science Book a Day

Synopsis: ‘I have a truly marvellous demonstration of this proposition which this margin is too narrow to contain.’

It was with these words, written in the 1630s, that Pierre de Fermat intrigued and infuriated the mathematics community. For over 350 years, proving Fermat’s Last Theorem was the most notorious unsolved mathematical problem, a puzzle whose basics most children could grasp but whose solution eluded the greatest minds in the world. In 1993, after years of secret toil, Englishman Andrew Wiles announced to an astounded audience that he had cracked Fermat’s Last Theorem. He had no idea of the nightmare that lay ahead.

In ‘Fermat’s Last Theorem’ Simon Singh has crafted a remarkable tale of intellectual endeavour spanning three centuries, and a moving testament to the obsession, sacrifice and extraordinary determination of Andrew Wiles: one man against all the odds.

First Published: 1997, Reissued: 2002| ISBN-13: 978-1841157917

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## 7 thoughts on “Fermat’s Last Theorem”

1. Pingback: Mathematics
2. Pham Duc Sinh says:

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

Xp + Yp ?= Zp (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), (9), 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 of 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?

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1. mathtuition88 says:

Yes, it is interesting. Thanks for commenting!

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3. Pham Duc Sinh says:

I am sorry that the typing does not reflect X.X…X (p times) =X^p.
The typing corrected should be:
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), (9), 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 of 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?

Like

4. Pham Duc Sinh says:

I am sorry that the typing does not reflect X.X…X (p times) =X^p.
The corrected typing should be:
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), (9), 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 of 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?

Like

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