3.2. Fermat’s Pythagorean triangle

This proposition is the only one of which Fermat wrote the explicit demonstration by infinite descent and it has become a sort of prototype for understanding the method by Fermat. Many scholars dealth with it^{1}. Almost all reconstructions are similar, being based on the words by Fermat. Here we will follow the reconstruction by Mahoney’s^{2} adding some specifications if necessary. But at the end of the proof, we will add some considerations which are present neither in Mahoney’s nor in other reconstructions and that, in our opinion, are necessary to fully understand the method by Fermat.

Fermat wrote in his Observations sur Diophante, (observation 45):

Si area trianguli esset quadratus, darentur duo quadratoquadrati [that is two fourth powers] quorum differentia esset quadratus; [4]; unde sequitur dari duo quadratos quorum et summa et differentia esset quadratus [5]. Datur itaque numerus, compositus ex quadrato et duplo quadrati, aequalis quadrato, ea conditione ut quadrati eum componentes faciant quadratum. Sed, si numerus quadratus componitur ex quadrato et duplo alterius quadrati, eius latus similiter componitur ex quadrato et duplo quadrati [6], [7], ut facillime possumus demonstrare; unde concludetur latus illud esse summam laterum circa rectum trianguli rectanguli, et unum ex quadratis illud componentibus efficere basem, et duplum quadratum aequari perpendiculo.

Illud itaque triangulum rectangulum conficietur a duobus quadratis quorum summa et differentia erunt quadrati. At isti duo quadrati minores probabuntur primis quadratis primo suppositis, quorum tam summa quam differentia faciunt quadratum: ergo, si dentur duo quadrati quorum summa et differentia faciant quadratum, dabitur in integris summa duorum quadratorum eiusdem naturae, priore minor.

Eodem ratiocinio dabitur et minor ista inventa per viam prioris, et semper in infinitum minores invenientur numeri in integris idem praestantes. Quod impossibile est, quia, dato numero quovis integro, non possunt dari infiniti in integris illo minores^{3}.

Translation:

If the area of a triangle were a square, then would be given two quadrato-quadrates [fourth powers] of which the difference were a square [4]. Whence it follows that two squares would be given of which the sum and the difference would be squares [5]. And thus a number composed of a square and the double of a square would be given equal to a square, under the condition that the squares composing it make a square. But if a square number is composed of a square and the double of another square, its root is similarly composed of a square and the double of a square [6], [7], as we can most easily demonstrate. Whence one concludes that this root is the sum of the sides about the right angle, and that one of the squares composing it constitutes the base and that the double square is equal to the perpendicular.

1 A complete book concerning this theorem is Goldstein, 1995. The author presents the different reconstructions based on Fermat’s words. On the other hand, Frenicle gave a demonstration by descent of this theorem which is different from Fermat’s some aspects. Goldstein aims to underline the independence of Frenicle’s demonstration (Proposition XXXI of Frenicle 1675, 1729) from Fermat’s, while generally in the previous literature, there was the idea that Frenicle’s proof was actually the same as Fermat’s with some slight variations. For example Weil, 1984, p. 77, writes: “Frenicle follows this proof faithfully, with little more than verbal changes.”

Goldstein also reports and comments on demonstrations which are not connected to the indefinite descent. One also finds demonstrations of this theorem based on Fermat’s words, for instance, in Edwards, 1977, Paragraph 1.6. and in Weil, 1984, Chapter 2, Paragraph X. In order to clarify the logic of the descent, all the reconstructions of this theorem are equivalent, even if they are different in some mathematical details. For this reason, I will not enter other reconstructions. The reader can consult the book by Goldstein on this subject.

2 Mahoney, 1994, pp. 352-354. The paragraph’s passages printed in italics are mine.

3 Fermat, Oeuvres, 1, p. 340.

Hence, this right triangle is composed of two squares of which the sum and difference are squares. But these two squares will be proved to be smaller than the first squares initially posited, of which the sum and difference also made squares. Therefore, if two squares are given of which the sum and the difference are squares, there exists in integers the sum of two squares of the same nature, less than the former.

By the same argument there will be given in the prior manner another one less than this, and smaller numbers will be found indefinitely having the same property. Which is impossible, because, given any integer, one cannot give an infinite number of integers less than it^{4}.

So we have:

Thesis: there is no Pythagorean triangle, of which the area is equal to the square of an integer number.

1. The sides of a Pythagorean triangle can be written in the form:

p^{2} + q^{2}; 2pq; p^{2} − q^{2}

The first number represents the hypotenuse and the other two numbers the two small sides of the right triangle^{5}.

2. Hence the area of the triangle is

pq(p^{2} − q^{2})

We must prove that this number can never be the square of an integer. We assume that p and q are mutually prime and that the expression in 1. is a primitive triple, that is p^{2} + q^{2}; 2pq; p^{2} − q^{2} are mutually prime too (if the theorem is true for primitive triples, then it is also true for derived triples). Hence p and q are one even and the other odd and p > q .

3. It is possible to prove that, if p and q are mutually prime, then pq is prime with p^{2} − q^{2} . So the expression pq(p^{2} − q^{2}) is a square if and only if p,q and p^{2} − q^{2} are squares. Then

pq = b^{2}; p = d^{2}; q = f^{2}; p^{2} − q^{2} = c^{2}

1. Thus, we have:

a^{2} + 2b^{2} .

4 This translation is by Mahoney, in Mahoney, 1994, pp. 352-353.

5 It is clear that this condition is sufficient in order to have a Pythagorean triple. It is not difficult to prove that the condition is necessary too (Hardy-Wright, 2000, pp. 190-191). With a similar argument it is possible to demonstrate that, as Fermat says, the equation z^{2} = x^{2} + 2y^{2} is solvable if and only if z is of the form

d^{4} − f^{4} = c^{2}

And these are the two quadrato-quadrati (that is, two fourth powers), of which the difference is a square.

2. Now we can write the area of the triangle like this:

d^{2}f^{2}(d^{4} − f^{4}) = d^{2}f^{2}(d^{2} + f^{2})(d^{2} − f^{2})

with the condition that

{d^{2} + f^{2} = g^{2}? {d^{2} − f^{2} = h^{2}? no{?

Therefore, as Fermat writes, we have two squares, of which the sum and the difference is a square.

3. From the second equation of the system, we have

d^{2} = h^{2} + f^{2}

and replacing this value in the first equation, one obtains

g^{2} = h^{2} + 2f^{2}

7. But, as Fermat asserts, if one square is composed of one square and of the double of another square, then its root is composed of one square and of the double of another square too. Thus, it is possible to write:

g = k^{2} + 2m^{2}

8. Then

g^{2} = (k^{2} + 2m^{2})^{2} = k^{4} + 4k^{2}m^{2} + 4m^{4}

If one adds and subtracts the number 8k^{2}m^{2} to this expression, the following results

g^{2} = (k^{2} − 2m^{2}) + 2(2mk)^{2}

But

g^{2} = h^{2} + 2f^{2}

then it is possible to pose

f^{2} = 4k^{2}m^{2}

9. Finally we get to:

g^{2} = d^{2} + f^{2} (Point 5)
g^{2} = k^{4} + 4k^{2}m^{2} + 4m^{4} (Point 8)

d^{2} + f^{2} = k^{4} + 4k^{2}m^{2} + 4m^{4} f^{2} = 4k^{2}m^{2} (Point 8)

So that

d^{2} = k^{4} + 4m^{4}

10. Therefore d, k^{2} and 2m^{2} are a new Pythagorean triple and hence they can be considered respectively as the hypotenuse and the two small sides of another right triangle, of which the area is the square k^{2}m^{2} .

This is Mahoney’s reconstruction with a few explanations added. Now we will follow an argumentation that is different and a little more specific than Mahoney’s.

In order to conclude the proof, it is necessary to show that: 1) the argument can be iterated; 2) the hypotenuse (or another dimension) of the derived triangle is smaller than the hypotenuse (or of the corresponding dimension) of the first triangle. Namely, it is necessary to specify the relation between the first and the derived triangle.

Since d, k^{2} and 2m^{2} represent a new Pythagorean triple, they are respectively of the forms:

d = p^{'2} + q^{'2};

k^{2} = p^{'2} − q^{'2}; 2m^{2} = 2p'q'

Consequently the area is

k^{2}m^{2} = p'q'(p^{'2} − q^{'2})

Therefore, applying the same reasoning for the first triangle, one sees that p' and q' are squares, and hence it is possible to put

p' = u^{2}; q' = v^{2}

The hypotenuse d is such that

d = u^{4} + v^{4}

Let us remember that the hypotenuse of the first triangle is equal to d^{4} + f^{4} . Therefore all the measurements (hypotenuse, small sides, and area) of the two triangles can be expressed in the same form and consequently, the process can be iterated indefinitely.

But, as we have seen, the hypotenuse of the first triangle is equal to

p^{2} + q^{2}
and p = d^{2} and q = f^{2}

so that this hypotenuse is equal to

d^{4} + f^{4}

But

d <? d^{4} + f^{4} .

This last inequality is surely true because d = \sqrt{p} , but the hypotenuse of the first triangle is p^{2} + q^{2} and p>1, so that \sqrt{p} > 1 . Besides, since \sqrt{p} = d = p^{'2} + q^{'2} > 1 , and p' > q' , then p' > 1 (let us remember that every letter denotes a positive integer). Furthermore, if we construct a third triangle, its hypotenuse will be equal to \sqrt{p'} > 1 and so on. Therefore, this argumentation shows that it is impossible to arrive at the triple 1^{2} + 0^{2} = 1^{2} . In fact, this triple is certainly not applicable to any triangle (one side is 0), but the hypothetic area would be 0, and 0 is a square. Consequently, if the descent reached the triple 1^{2} + 0^{2} = 1^{2} , and, if we repeated the process backward, we would obtain triples that are applicable to Pythagorean triangles and these triangles would have the area equal to the square of an integer number. From a certain point of view, this process would have some analogies with the one used in order to solve Pell’s equation.

The “smallest” Pythagorean triple is (1,0,1). Given a triple (a,b,c) such that \frac{ab}{2} = x^{2} ,

the process described allows us to reach a triple (a',b',c') , such that \frac{a'b'}{2} = x'^{2} , being c' < c . The procedure can be iterated and since we remain in the field of the positive integers, it would be necessary to find a value c_{n} = 1 , but this is impossible. Therefore we are in the absurd situation of having to admit the existence of infinite integers between c and 1. The indefinite descent is just this.

From this demonstration, it is possible to deduce one of the most important properties inherent in every argument by indefinite descent: there is an invariable form with different orders of sizes and that represents an integer number. The existence of this form guarantees the iterability of the process. In this case, such a form is provided by the sum of the fourth powers that constitutes the hypotenuses.

Furthermore, we underline the great logical complexity of the demonstration: first of all, from a mathematical point of view, it is not easy to construct the descent and to show that there is the invariable form. But also when one has constructed these steps, other logical considerations are necessary, as seen above. In fact, the existence of the invariable form with different orders of size is a necessary condition, but not a sufficient one for a demonstration by descent.

It seems to me that this proposition concerning the Pythagorean triangle in numbers is not only important in itself, but also because it represents a sort of will by which Fermat has exposed a method, the indefinite descent. In a sense, the demonstration of this theorem is difficult enough to show the power of the method, but easy enough to clearly explain the main properties of the descent without excessive mathematical complications.