33. To find two numbers in a given ratio and such that the difference of their squares also has to their sum a given ratio.

Given ratios 3:1 and 6:1.

Lesser number x, which is found to be 3.
The numbers are 3, 9.

34. To find two numbers in a given ratio and such that the difference of their squares also has to their difference a given ratio.

Given ratios 3:1 and 12:1.

Lesser number x, which is found to be 3.

The numbers are 3, 9.

Similarly by the same method can be found two numbers in a given ratio and (1) such that their product is to their sum in a given ratio, or (2) such that their product is to their difference in a given ratio.

35. To find two numbers in a given ratio and such that the square of the lesser also has to the greater a given ratio. Given ratios 3:1 and 6:1 respectively. Lesser number x, which is found to be 18. The numbers are 18, 54.

36. To find two numbers in a given ratio and such that the square of the lesser also has to the lesser itself a given ratio.

Given ratios 3:1 and 6:1.

Lesser number x, which is found to be 6.
The numbers are 6, 18.

37. To find two numbers in a given ratio and such that the square of the lesser also has to the sum of both a given ratio.

Given ratios 3:1 and 2:1.

Lesser number x, which is found to be 8.

The numbers are 8, 24.

38. To find two numbers in a given ratio and such that the square of the lesser also has to the difference between them a given ratio.

Given ratios 3:1 and 6:1.

Lesser number x, which is found to be 12.

The numbers are 12, 36.

:

1

Similarly can be found two numbers in a given ratio and

(1) such that the square of the greater also has to the lesser a given ratio, or

(2) such that the square of the greater also has to the greater itself a given ratio, or

(3) such that the square of the greater also has to the sum or difference of the two a given ratio.

39. Given two numbers, to find a third such that the sums of the several pairs multiplied by the corresponding third number give three numbers in arithmetical progression.

Given numbers 3, 5.

Required number x.

:

The three products are therefore 3x + 15, 5x + 15. 8x.
Now 3x+15 must be either the middle or the least of

the three, and 5x + 15 either the greatest or the
middle.

(1) 5x+15 greatest, 3x + 15 least.

Therefore 5x + 15 + 3x + 15-2.8x, and

(2) 5x+15 greatest, 3x + 15 middle.

Therefore (5x+15)-(3x+15)=3x+15-8t, and

[The first five problems of this Book are mere repetitions of problems in Book I. They probably found their way into the text from some ancient commentary. In each case the ratio of one required number to the other is assumed to be 2:1. The enunciations only are here given.]

1. To find two numbers such that their sum is to the sum of their squares in a given ratio [cf. 1. 31].

2. To find two numbers such that their difference is to the difference of their squares in a given ratio [cf. I. 34].

3. To find two numbers such that their product is to their sum or their difference in a given ratio [cf. I. 34].

4. To find two numbers such that the sum of their squares is to their difference in a given ratio [cf. I. 32].

5. To find two numbers such that the difference of their squares is to their sum in a given ratio [cf. I. 33].

61. To find two numbers having a given difference and such that the difference of their squares exceeds their difference by a given number.

Necessary condition. The square of their difference must be less than the sum of the said difference and the given excess of the difference of the squares over the difference of the numbers.

Difference of numbers 2, the other given number 20.
Lesser number x. Therefore x + 2 is the greater, and
4x + 4 = 22.

Therefore x = 44, and

the numbers are 44, 64.

71. To find two numbers such that the difference of their squares is greater by a given number than a given ratio of their difference. [Difference assumed.]

Necessary condition. The given ratio being 3:1, the square of the difference of the numbers must be less than the sum of three times that difference and the given number.

Given number 10, difference of required numbers 2.
Lesser number x. Therefore the greater is x + 2, and

4x+43.2 + 10.

Therefore x = 3, and

the numbers are 3, 5.

8. To divide a given square number into two squares.

The problems 11. 6, 7 also are considered by Tannery to be interpolated from some ancient commentary.

• Here we have the identical phrase used in Euclid's Data (cf. note on p. 133 above) : the difference of the squares is τῆς ὑπεροχῆς αὐτῶν δοθέντι ἀριθμῷ μείζων ἢ ἐν λόγῳ, literally "greater than their difference by a given number (more) than in a (given) ratio," by which is meant "greater by a given number than a given proportion or fraction of their difference."

• It is to this proposition that Fermat appended his famous note in which he enunciates what is known as the "great theorem" of Fermat. The text of the note is as follows:

"On the other hand it is impossible to separate a cube into two cubes, or a

•

Given square number 16.

x' one of the required squares. Therefore 16-x must be equal to a square.

Take a square of the form1 (mx-4), m being any integer and 4 the number which is the square root of 16, eg. take (2x-4), and equate it to 16-x Therefore 4x - 16x + 16-16-x',

or 5x = 16x, and x = 8.

The required squares are therefore

256 144

25' 25

9. To divide a given number which is the sum of two squares into two other squares.

biquadrate into two biquadrates, or generally any power except a square into two porvers with the same exponent. I have discovered a truly marvellous proof of this, which however the margin is not large enough to contain."

Did Fermat really possess a proof of the general proposition that x+y=s** cannot be solved in rational numbers where m is any number >a? As Wertheim says, one is tempted to doubt this, seeing that, in spite of the labours of Euler, Lejeune-Dirichlet, Kummer and others, a general proof has not even yet been discovered. Euler proved the theorem for m=3 and m=4, Dirichlet for m = 5, and Kummer, by means of the higher theory of numbers, produced a proof which only excludes certain particular values of m, which values are rare, at all events among the smaller values of m; thus there is no value of m below 100 for which Kummer's proof does not serve. (I take these facts from Weber and Wellstein's Encyclopädie der Elementar-Mathematik, 19, p. 284, where a proof of the formula for m=4 is given.)

It appears that the Göttingen Academy of Sciences has recently awarded a prize to Dr A. Wieferich, of Münster, for a proof that the equation x+y=s cannot be solved in terms of positive integers not multiples of p, if 2 - 2 is not divisible by p. "This surprisingly simple result represents the first advance, since the time of Kummer, in the proof of the last Fermat theorem" (Bulletin of the American Mathematical Society, February 1910).

Fermat says ("Relation des nouvelles découvertes en la science des nombres," August 1659, Oeuvres, II. p. 433) that he proved that no cube is divisible into two cubes by a variety of his method of infinite diminution (descente infinie or indisfinic) different from that which he employed for other negative or positive theorems; as to the other cases, see Supplement, sections I., 11.

1 Diophantus' words are: "I form the square from any number of ἀριθμοί minus as many units as there are in the side of 16." It is implied throughout that m must be so chosen that the result may be rational in Diophantus' sense, i.e. rational and positive.

• Diophantus' solution is substantially the same as Euler's (Algebra, tr. Hewlett, Part II. Art. 219), though the latter is expressed more generally.

Given number 13 = 2*+-3.

As the roots of these squares are 2, 3, take (x + 2) as the first square and (mx - 3) as the second (where m is an integer), say (2x - 3).

Therefore (x2 + 4x + 4) + (4x2 + 9 - 12x) = 13,

10. To find two square numbers having a given difference.
Given difference 60.

Side of one number x, side of the other x plus any
number the square of which is not greater than 60,
say 3.

the required squares are 721, 1321.

11. To add the same (required) number to two given numbers

so as to make each of them a square.

(1) Given numbers 2, 3; required number x.

This is called a double-equation (διπλοῖσότης).

To solve it, take the difference between the two expressions and resolve it into two factors'; in this case let us say 4.

Then take either

(a) the square of half the difference between these factors and equate it to the lesser expression,

or (b) the square of half the sum and equate it to the