The proof itself was found after his death in his notes on Diophantus's Arithmetica.

In what follows, I will go through each sentence of Fermat's proof and provide details to make his step clear. My goal is to shed light on Fermat's thinking in his one existent proof.

I will show how this proof can also be used to prove the case of Fermat's Last Theorem where n = 4. In my next blog, I go over a simpler proof for n=4.

My analysis is based on the work by Harold M. Edwards called Fermat's Last Theorem. The translation is taken from T. L. Heath's translation.

Fermat writes:

(1) Fermat: "If the area of a right-angled triangle were a square, there would exist two biquadrates the difference of which would be a square number."

A biquadrate is a value to the fourth-power. So, the biquadrate of

**2**is

**2**.

^{4}= 16This corresponds to the following equation:

**p**.

^{4}- q^{4}= z^{2}When I first read this, it was surprising to me that Fermat assumed that this equation was obvious from the problem of showing that right triangle's area cannot be equal to a square.

This equation proves n = 4 since if

**x**, then:

^{4}+ y^{4}= z^{4}**x**

^{4}= (x^{2})^{2}= z^{4}- y^{4}So, proving there is no right triangle that has an area equal to a square will also prove Fermat's Last Theorem for n=4.

The steps to this equation can be traced as follows:

(a) A right triangle is characterized by the Pythagorean Theorem:a^{2}+ b^{2}= c^{2}.

(b) The area of a rectangle is base x height.

(c) A right triangle of a given base and height is created by dividing the same rectangle across the diagonal.

(d) Therefore, the area of a right triangle is base * height / 2.

(e) Or, usinga,b,cfrom above: area= ab/2

(f) From the solution to Pythagorean Triples, we know that:a = (2pq)db = (p^{2}- q^{2})d

where we know that p,q are relatively prime.

(g)So, saying the area of a right triangle is equal to a square comes down to proving that there are no solutions for:z^{2}= ((2pq)d[p^{2}- q^{2}]d)/2 = (pq)(d^{2})[p^{2}- q^{2}]

(h) From a previous result, we know thatdwill dividezso, we are left with showing no solutions for:(z/d)^{2}= (pq)[p^{2}- q^{2}]

(i) Sincep,qare relatively prime, it follows that(pq)and[pare relatively prime [see here for details]^{2}- q^{2}]

(j) From this we conclude, thatpqis a square andpare squares.^{2}- q^{2}

(k) And sincep,qare relatively prime,p,qare themselves squares.

(l) So, there existP,Qsuch thatp = Pand^{2}q = Q(m) Since^{2}pqis a square, there exists a valueksuch thatk.^{2}= pq

(n) And from (m),kdivides(z/d), so we get:(z/dk)^{2}= p^{2}- q^{2}= P^{4}- Q^{4}

^{}(2) Fermat: "Consequently there would exist two square numbers the sum and difference of which would both be squares."

This one is a bit easier to derive

**z**

^{2}= P^{4}- Q^{4}= (P^{2}+ Q^{2})(P^{2}- Q^{2})Now, all, we need to show is that

**(P**is relatively prime to

^{2}+ Q^{2})**(P**which means that

^{2}- Q^{2})**p + q**is realtively prime to

**p - q**. [ The trick is to remember that p,q are relatively prime and they are of different parity (see 13(b) here). See here for details]

In other words, there exist two square numbers (P

^{2},Q

^{2}

**) the sum (**

^{}**P**) and difference (

^{2}+ Q^{2}**P**) are both squares.

^{2}- Q^{2}(3) Fermat: "Therefore we should have a square number which would be equal to the sum of a square and the double of another square..."

We know that

**P**is equal to a square. Let's say

^{2}+ Q^{2}**S**.

^{2}We also know that

**P**is equal to a square. Let's say

^{2}- Q^{2}**T**

^{2}So that

**P**

^{2}= Q^{2}+ T^{2}Which combined with the other equation gives us:

**Q**

^{2}+ T^{2}+ Q^{2}= T^{2}+ 2Q^{2}= S^{2}We can assume that

**S,P, Q**are relatively prime and also that

**P,T,Q**are relatively prime with

**S,P,T**odd and

**Q**even. [See here for details]

We can also assume that

**Q,S,T**are relatively prime. [See here for details]

(4) Fermat: "...while the squares of which this sum is made up would themselves have a square number for their sum."

And as stated before:

**P**.

^{2}= Q^{2}+ T^{2}(5) Fermat: "But if a square is made up of a square and the double of another square, its side, as I can very easily prove, is also similarly made up of a square and the double of another square."

So, he is saying this:

**T**--> S =

^{2}+ 2Q^{2}= S^{2}**t**.

^{2}+ 2q^{2}This means that

**2Q**

^{2}= S^{2}- T^{2}= (S - T)(S + T)And:

**Q**

^{2}= (1/2)(S-T)(S+T)Now

**Q,S,T**are relatively prime. We know that

**Q, S-T, S+T**are even. [see above for details]

Let

**S-T = 2u**,

**S + T = 2v**

So that

**Q**

^{2}= (1/2)(2u)(2v) = 2uvNow,

**u,v**are relatively prime [see here for details]

And, either

**u**or

**v**is even, let's assume

**u**

So that,

**u = 2w**and

**Q**

^{2}= 2(2w)vAnd

**w,v**are relatively prime so

**w,v**are squares.

Let

**w = W**,

^{2}**v = V**

^{2}And

**S + T + S - T = 2S = 2u + 2v = 2(2W**

^{2}+ V^{2})So that

**S = 2W**

^{2}+ V^{2}(6) Fermat: "From this we conclude that the said side is the sum of the sides about the right angle in a right-angled triangle and that the simple square contained in the sum is the base and the double of the other square is the perpendicular."

So, he is saying this:

**S =**

**2W**

^{2 }**+ V**-> (

^{2}**2W**)

^{2}^{2}+ (

**V**)

^{2}**is a square.**

^{2}(

**2W**)

^{2}^{2}+ (

**V**)

^{2}**= (**

^{2 }**2w**)

^{2}+ (

**v**)

**= (**

^{2 }**u**)

^{2}+ (

**v**)

**=**

^{2}[(1/2)(S-T)]

^{2}+ [(1/2)(S+T)]

^{2}=

**[S**

^{2}- 2ST + T^{2}]/4 + [S^{2}+ 2ST + T^{2}]/4 =**= [2S**

^{2}+ 2T^{2}]/4 = (S^{2}+ T^{2})/2.Now, since

**S**

^{2}= 2Q^{2}+ T^{2}We have:

**(2Q**

^{2}+ T^{2}+ T^{2})/2 = Q^{2}+ T^{2}Which equals

**P**.

^{2}(7) Fermat: "This right-angled triangle will thus be formed from two squares, the sum and differences of which will be squares."

We can once again apply the Solution to the Pythagorean Triples so that:

**2W**=

^{2}**2mn**

**V**=

^{2}**m**

^{2}- n^{2}**P**=

**m**

^{2}+ n^{2}So we've proven that the difference is a square. Now, we need to prove that the sum is a square.

Since

**Q**= 4W

^{2}= 2(2w)v^{2}V

^{2}.

**Q**= 2(

^{2}**2mn**)(

**m**)=4mn(

^{2}- n^{2}**m**).

^{2}- n^{2}And

**(Q/2)**

^{2}= mn(m^{2}- n^{2})Since

**m,n**and

**m**are relatively prime, then

^{2}- n^{2}**m,n,m**are all squares.

^{2}-n^{2}Letting m = M

^{2}and

**n = N**

^{2}We get:

**V**

^{2}= M^{4}- N^{4}= (M^{2}- N^{2})(M^{2}+ N^{2})Since

**M**are relatively prime, both the sum and differences are squares.

^{2}- N^{2}, M^{2}+ N^{2}(8) Fermat: "But both these squares can be shown to be smaller than the squares originally assumed to be such that both their sum and differences are squares."

So M

^{2}+ N

^{2}≤ P

^{2}+ Q

^{2}.

Since M

^{2}+ N

^{2}≤ m + n is less than

**P**which is less than

**P**.

^{2}+ Q^{2}(9) Fermat: "Thus if there exist two squares such that their sum and differences are both squares, there will also exist two other integer squares which have the same property but a smaller sum"

We have proven that:

if

**P**is a square and

^{2}+ Q^{2}**P**is a square, then there exists

^{2}- Q^{2}**M,N**such that:

(a) M

^{2}+N

^{2}is less than

**P**

^{2}+ Q^{2}(b)

**M**is a square

^{2}+ N^{2}(c)

**M**is a square.

^{2}- N^{2}(10) Fermat: "By the same reasoning we find a sum still smaller than the last found, and we can go on ad infinitum finding integer square numbers smaller and smaller which have the same property."

This leads to an infinite number of smaller solutions.

(11) Fermat: "This is, however, impossible because there cannot be an infinite series of numbers smaller than any given integer we please."

So that we have a proof by Infinite Descent.

(12) Fermat: "The margin is too small to enable me to give the proof completely and with all detail."

At least, this time we were able to reconstruct the proof. :-)

-Larry