Pierre de Fermat was a French lawyer and mathematician born in 1607...

Note that one of $a$, $b$ has to be odd and the other one even. If ...

If $y=n^2$ and $z=m^2$ it would be possible to construct a triangle...

- If $y$ is a square then, $z=a^2+b^2=(\sqrt{z})^2$ which is just t...

Let's assume we have a triangle with sides $A^2$, $B^2$ and hypoten...

Euclid's formula is a fundamental formula for generating Pythagorea...

The observation in the concluding paragraph is recognisable to stud...

Fermat’s method of "descente inﬁnie" - Proof of Fermat’s Last Theorem for n=4

Stan Dolan

126A Harpenden Road, St. Albans

Introduction

Fermat proved the impossibility of ﬁnding positive in-

teger solutions of the equation x

4

+ y

4

= z

4

by a method

which he claimed would ’enable extraordinary develop-

ments to be made in the theory of numbers’ [1], His

method was based upon the idea of using a supposed

solution to ﬁnd another solution, smaller than the ﬁrst.

Proceeding in this fashion he could therefore obtain an

inﬁnite series of smaller and smaller solutions. At this

point he concluded his proof as follows:

’This is, however, impossible because there cannot be

an inﬁnite series of numbers [positive integers] smaller

than any given [positive] integer we please. The margin

is too small to enable me to give the proof completely and

with all details [1].’

The margin referred to here was in Fermat’s copy of

the Arithmetica, an ancient Greek text written by Dio-

phantus of Alexandria. The problems given in this text

provided the stimulus for many of Fermat’s discoveries

in number theory. The equation x

4

+ y

4

= z

4

arose from

problems concerning right-angled triangles

Fermat’s proof

Fermat gave suﬃcient details of his proof to enable it

to be reconstructed. However, the purpose of this note

is not to reproduce the original proof faithfully but to

modify and simplify it whilst retaining the historical con-

nection with right-angled triangles.

Theorem: A right-angled triangle with rational sides

cannot have two sides each with a length equal to a square

or twice a square.

Proof. Suppose such a triangle did exist. By scaling

by a square number we can further assume the sides to

have integer lengths. Then, any prime factor p of two of

the lengths would be a factor of all three lengths. Fur-

thermore, if p were odd, then p

2

would be a factor of two

of the lengths and thus of all the lengths. By cancelling

down as necessary, we can therefore assume the lengths

to be coprime in pairs. Then, by Euclid’s formula, there

are coprime positive integers a and b, one odd and one

even, such that the triangle is as shown.

FIG. 1:

Now y and z are both odd and so neither can be twice

a square. If they were both squares then the triangle

shown below would be a new solution.

FIG. 2:

Otherwise, just one of y and z is a square and x = 2ab

is a square or twice a square. Then each of a and b is a

square or twice a square and one of the triangles shown

below would be a new solution.

FIG. 3:

In all cases, the new solution has a smaller hypotenuse

than the original triangle. Since there cannot be an in-

ﬁnite series of decreasing hypotenuses, there can be no

triangle with the required properties.

2

Concluding Remarks

A positive rational solution of any of the equations

x

4

+ y

4

= z

4

x

4

− y

4

= z

4

x

4

+ 4y

4

= z

4

x

4

− 4y

4

= z

4

(1)

would give rise to a right-angled triangle contradict-

ing the Theorem. All of these equations are therefore

insoluble.

It is worth noting that it is not obvious how to use a

supposed solution of x

4

+ y

4

= z

4

to ﬁnd another solu-

tion of the same equation. However, by strengthening

the conjecture as shown above it was possible to carry

out the proof relatively simply. This illustrates an im-

portant feature of Fermat’s method of proof; that the

choice of conjecture can be crucial to the success of the

method. Sometimes it can be much easier to prove a

stronger result than the one which is needed.

[1] Sir Thomas L. Heath, Diophantus of Alexandria, Dover,

1964, p. 293.

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We can repeat this operation which will lead to an infinite series of triangles with decreasing hypotenuses, thus not being possible.
```
I think the proof is not yet making this conclusion at this paragraph yet, it's merely generalizing the case where the sides have a common prime factor under the case where the sides are coprimes. Hence the following sentence:
```
By cancelling down as necessary, we can therefore assume the lengths to be coprime in pairs.
```
If $y=n^2$ and $z=m^2$ it would be possible to construct a triangle with the length of the hypotenuse equal to $\frac{y+z}{2} = \frac{a^2+b^2+a^2-b^2}{2}= a ^2 = \frac{n^2 + m^2}{2} $ and the other sides equal to $\frac{z-y}{2}= \frac{a^2+b^2-a^2+b^2}{2}= b^2 = \frac{m^2 - n^2}{2}$ and $2\frac{m}{\sqrt{2}}\frac{n}{\sqrt{2}} = \sqrt{yz}$. Note that the new sides follow the structure defined by Euclid's formula.
Euclid's formula is a fundamental formula for generating Pythagorean triples (three positive integers a, b, and c, such that $a^2 + b^2 = c^2$) given an arbitrary pair of integers $m$ and $n$ with $m > n > 0$. The formula states that the integers
$$
a =m^{2}-n^{2} \\
b =2mn \\
c =m^{2}+n^{2}
$$
form a Pythagorean triple. The triple generated by Euclid's formula is primitive - one in which $a, b$ and $c$ are coprime - if and only if $m$ and $n$ are coprime and not both odd.
Let's assume we have a triangle with sides $A^2$, $B^2$ and hypotenuse C.
If 2 of the sides, for instance $A^2$ and $C$, have a common prime factor $p$, then
$A^2=K_Ap$ and $C=K_Cp$
where $K_A,K_C$ are just the product of the rest of the prime factors.
Using the Pythagorean Theorem
$$
(A^2)^2+(B^2)^2=C^2
$$
and so $B^2=\sqrt{C^2-(A^2)^2}=\sqrt{K_C^2-K_A^2}p$ is also divisible by $p$. This way, it's always possible to divide all sides by $p$ getting a smaller triangle with integer lengths. We can repeat this operation which will lead to an infinite series of triangles with decreasing hypotenuses, thus not being possible.
- If $y$ is a square then, $z=a^2+b^2=(\sqrt{z})^2$ which is just the Pythagorean theorem for the triangle on the right (Figure 3).
- If $z$ is a square then, $y = a^2-b^2 \equiv a^2 = (\sqrt{y})^2+b^2 $ which is just the Pythagorean theorem for the triangle on the left (Figure 3).
Note that one of $a$, $b$ has to be odd and the other one even. If $a$ is odd and $b$ is even, then $a^2$ is also odd and $b^2$ is even. Then both $a^2+b^2 = z$ and $a^2-b^2 = y$ are also odd, since odd +/- even = odd.
Pierre de Fermat was a French lawyer and mathematician born in 1607 who is given credit for his principle for light propagation and his Last Theorem.
In number theory, Fermat's Last Theorem (FLT) states that no three positive integers $a, b$ and $c$ satisfy the equation $a^n + b^n = c^n$ for any integer value of $n$ greater than 2.
Fermat claimed in 1637 in the margin of a copy of Arithmetica that he had a proof for FLT that was too large to fit in the margin. It took however 358 years of effort by some of the most talented mathematicians before Andrew Wiles released the first successful proof in 1994.
In the meantime there were several proofs for specific exponents, including the one for $n=4$ which is particularly interesting because it was the only complete proof left by Pierre de Fermat.
![](https://www.thefamouspeople.com/profiles/images/og-pierre-de-fermat-5067.jpg)
The observation in the concluding paragraph is recognisable to students of mathematics.
The method of ‘infinite descent’ is essentially a proof by induction, relying as it does on the well-foundedness of the natural numbers. Students will be familiar with the need to have a strong enough induction hypothesis to prove what is often a weaker statement. This corresponds to the ‘choice of conjecture’ of the article.