Diophantus of Alexandria

By means of the above two lemmas combined we can now prove that the area of a primitive right-angled triangle cannot be a square number.

Let the sides of the triangle be aa + b2, a3 – b3, zab. If now the area were square, the product of the perpendicular sides would be double of a square. But the perpendicular sides are prime to one another. There fore the odd perpendicular a2 - would be a square, and the even perpendicular zab the double of a square. But, if a- were equal to , we could (by the first Lemma) find a second primitive triangle with smaller sides in which the odd perpendicular would be, the even per pendicular, and the hypotenuse a. Again, since zab would be double of a square, ab would be a square, and, since a is prime to b, both a and b would be squares. The second triangle would accordingly have a square number both for its hypotenuse (a) and for its even perpendicular (8). That is, the second primitive triangle would satisfy the conditions of the second Lemma, and we could accordingly derive from the second primitive triangle a third primitive triangle with still smaller sides which would, exactly like the first triangle, have a square number for its odd perpendicular, and for its even perpendicular the double of a square number.

From this third triangle we could obtain a fourth, and by means of the fourth we could obtain a fifth with the same property as the first, and so we should have an unending series of primitive right-angled triangles, each successive triangle having smaller sides than the one before, and all being such that the odd perpendicular would be a square number, the even perpendicular the double of a square number, and consequently the area a square number. This, however, is impossible since there cannot be an unending series of integral numbers less than any given integral number.

Frénicle proves, by similar considerations, that neither can the area of a right-angled triangle in rational numbers be the double of a square number.

In enunciating Fermat's problems on right-angled triangles I shall in future for brevity and uniformity use &,, to denote the three sides, while will always represent the hypotenuse and E, the two perpendicular sides. 2. To find a right-angled triangle (C, E, n) such that

<= v3 &+n=v2

[Since +, this problem is equivalent to that of finding x, y such

that

x+y=v2 x2 + y2 = w

which is Question 17 in Chapter XIV. of Euler's Algebra, Part 11.]

First method.

Form a right-angled triangle from the numbers x + 1, x; the sides will

then be

(= 2x2 + 2x + 1, έ= 2x+1, n = 2x2 + 2x.

We have then the double-equation

2x2+ 2x + I = 162

2x2 + 4x + 1 = v2

The ordinary method of Diophantus gives the solution --; the riangle will therefore be formed from and 2 or, if we take the umerators only, – 5 and 12, and the triangle is (169, - 119, 120) which

s equally the result of forming a triangle from + 5 and + 12.

But, as one of the perpendiculars is negative, we must find another value of x which will make all three sides positive.

We accordingly form a triangle from x+5 and 12, instead of from 5 and 12, and repeat the operation. This gives for the sides

(= x2+10x+169, έ= x2+10x - 119, n=24x + 120,

and we have to solve the double-equation

x2+10x + 169 = u3,

x2 + 34x + I = v.

Making the absolute term the same in each, we have to solve

x2 + 10x + 169 = u3,

169x2 + 5746x + 169 = v2.

The difference is 168x2 + 5736x, which we may separate into the factors 14x, 12x + 2868 (the sum of the terms in x being 26x or 2.13x).

Equating the square of half the sum of these factors to the larger expression, or the square of half their difference to the smaller, we find in the usual way

The triangle is therefore formed from 21/0905

246792, and the triangle itself is

1888, 12, or from 2150905,

4687298610289, 4565486027761, 1061652293520,

the hypotenuse and the sum of the other two sides being severally squares.

Second method.

This is the same as the first method up to the forming of the triangle from x + 5 and 12 and the arrival at the double-equation

x2+10x+169 = u3,
x2+34x+ I = v2.

Multiply the two expressions together, and we must have

x + 44x3 + 510x2 + 5756x + 169 = a square

this gives, as a matter of fact, the same value of x, namely

In his note on Diophantus vi. 22 Fermat says that he confidently asserts that the above right-angled triangle is the smallest right-angled triangle in rational numbers which satisfies the conditions.

[The truth of this latter assertion was proved by Lagrange3. Lagrange observes that, since έ + n = y3, E2 + y2 = x^, say, we have, if we put z for έ-7,

and, if x, y is any solution of the latter equation,

= } (y2+ z), n = { (y2 — z).

For comparison we may give Euler's solution (Algebra, Part II., Art. 240; Commen tationes arithmeticae, II. p. 398).

First make x2+y3 a square by putting x=a2 – b2, y=2ab, so that

x8+ y2= (a2+62)3.

To make the last expression a fourth power put a=p3 - q3, b=2pg, so that

and accordingly

a2+b2=(p2+q2)2,
x2+ y2= (p2+q2)*.

In solving this we have to note that p, q should be positive, must be >q (for otherwise y would be negative), and a>b in order that x may be positive.

Put

and we obtain

But, if we put

+4p3q-6p3y3— 4pq3+q1 = (p3 −2pq+q2)3, 4P3q − 6p2q3= − 4p3q+6p3q2, whence plq=3.

=3, 9=2, we find x= 119, a negative value.

To find fresh values, we can substitute for the expression 9+ and solve for the ratio q/r; then, by taking for 9 the numerator and for r the denominator of the fraction so found, we find a value for and thence for x, y. This is Euler's method in the Algebra. But we avoid the necessity for clearing of fractions if (as in the Comment. arithm.) we leave a as the value of q and substitute 3+v for 3 as the value of p. We then have p1 = 81+1080+5422 + 1223 +24, +4p3q= 216+216v+72v2+ 8v3, -6p2q2=-216- 1440 – 2403,

- 4pq3= - 96- 32v,

+qt= 16,

x+y=1+148v+10222 + 2023 +z^=a square=(1+74′′ – v2)3, say;

whence
and we obtain

and

Taking integral values, we put p=1469, 9=84.

Therefore a=1385.1553=2150905, b=168. 1469 = 246792,

x=4565486027761, y=1061652293520,

which is the same as Fermat's solution.

• N. Mémoires de l'Acad. Royale des Sciences et Belles-lettres de Bérlin, année 1777, Berlin, 1779 Oeuvres de Lagrange, IV. pp: 377-398.

SUPPLEMENT

He sets himself therefore to find a general solution of the equation → and effects it by a method which is a variation of Fermat's 'scente, one of the most fruitful methods, as Lagrange observes, in the hole theory of numbers. The modified method consists of two parts, ) a proof that, assuming that there exist integral values of x, y greater lan I which satisfy the condition 2x-y=z, there are still smaller tegral values which will also satisfy it, (2) the discovery of a general ethod of deducing the latter from the former. This being done, and : being known that x=1, y = 1 are the minimum values, the successive igher values are found by reversing the process. Lagrange found that he four lowest values for x, y give the following pairs of values for έ, n, iamely

(1) έ= 1

, n=0,

, n=-119,

,n=-473304,

(4) = 1061652293520, n=4565486027761,

30 that the last pair (4) are in truth, as Fermat asserts, the smallest possible values in positive integers.]

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Solved in the ordinary way, this gives x=-1; consequently the triangle is formed from -, I, or from

5, 12.

We could proceed, as in the last problem, to deduce a new value for x, but we observe that the triangle formed from 5, 12, i.e. the triangle 169, 119, 120, satisfies the conditions.

4. To find a right-angled triangle L, E, n such that