Take half of 13 or 6, and we have to add to 6 a small fraction which will make it a square, I or, multiplying by 4, we have to make + 26 a square, i.e. 26x2 + 1 = a square =(5x+1), say, whence x = 10. That is, in order to make 26 a square, we must add, or, to make 6 a square, we must add, and zbo +64 = (§1)2. Therefore we must divide 13 into two squares such that their sides may be as nearly as possible equal to §. [This is the παρισότητος ἀγωγή described above, pp. 95–8.] Now 13=22+ 3. Therefore we seek two numbers such that 3 minus the first, so that the first = H. ਝੰਗ, and 2 plus the second, so that the second = . We write accordingly (11x+2)2, (3 −9x)2 for the required squares [substituting x for]. The sum = 202x2 - 10x+13 = = 13. Therefore x=8T, and the sides are 181, 18. Subtracting 6 from the squares of each, we have, as the parts of unity, 4843 5358 10201' 10201 10. To divide unity into two parts such that, if we add different given numbers to each, the results will be squares. Let the numbers1 be 2, 6 and let them and the unit be represented in the figure, where DA-2, AB=1, BE=6, and G is a point in AB so chosen that DG, GE are both squares. Now DE 9. = Therefore we have to divide 9 into two squares such that one of them lies between 2 and 3. Let the latter square be, so that the other is 9-12, where 32 > 2. Take two squares, one >2, the other <3 [the former 1 Loria (op. cit. p. 150n.), as well as Nesselmann, observes that Diophantus omits to state the necessary condition, namely that the sum of the two given numbers plus 1 must be the sum of two squares. Therefore, if we can make a lie between these, we shall solve the problem. We must have > and < . Hence, in making 9 a square, we must find > and < . Put 9-(3-mx), say, whence x = 6m/(m2 + 1). The first inequality gives 72m > 17m2 + 17; and 36217.17 1007, the square root of which' is not greater than 31; Similarly from the inequality 19m2 + 19>72m we find1 m 18. Let m=3. Therefore 9-(3- 3x), and x=. and the segments of I are 1438, 1371). 2809' 2809 11. To divide unity into three parts such that, if we add the same number to each of the parts, the results are all squares. Necessary condition'. The given number must not be 2 or any multiple of 8 increased by 2. Given number 3. Thus 10 is to be divided into three squares such that each > 3. Take of 10, or 33, and find r so that 9.x +3 may be a square, or 302 + 1 = a square =(5x+1)2, say. Therefore x = 2, x=4, 1/x2=†, and √ +3} = 121 = a square. Therefore we have to divide 10 into three squares each of which is as near as possible to ἀγωγή.] Now 10 = 3'+1'= the sum of the three squares 9, 18. . or (multiplying by 30) 90, 24, 18 with 55, we must 1 L.e. the integral part of the root is 31. The limits taken in each case are a fortiori limits as explained above, pp. 61-3. 1 See p. 61, ante. * See pp. 108-9, ante. [Since then = 3 − 38=4+3b=+], we put for the sides of the required numbers 3-35x, 31x+1, 37x+8. The sum of the squares = 3555x2 - 116x + 10 = 10. and this solves the problem. 12. To divide unity into three parts such that, if three different given numbers be added to the parts respectively, the results are all squares. Given numbers 2, 3, 4. Then I have to divide 10 into three squares such that the first > 2, the second > 3, and the third > 4. Let us add of unity to each, and we have to find three squares such that their sum is 10, while the first lies between 2, 24, the second between 3, 3, and the third between 4, 4. It is necessary, first, to divide 10 (the sum of two squares) into two squares one of which lies between 2,2; then, if we subtract 2 from the latter square, we have one of the required parts of unity. Next divide the other square into two squares, one of which lies between 3, 34; subtracting 3 from the latter square, we have the second of the required parts of unity. Similarly we can find the third part'. 1 Diophantus only thus briefly indicates the course of the solution. Wertheim solves the problem in detail after Diophantus' manner; and, as this is by no means too easy, I think it well to reproduce his solution. I. It is first necessary to divide 10 into two squares one of which lies between 2 and 3. We use the παρισότητος ἀγωγή. The first square must be in the neighbourhood of 2}; and we seek a small fraction which when added to 2 gives a square: in other words, we must make 4 ( 2+ + 2) a square. This expression may be written 10+ -(-1); ;)*, and, to make this a square, we put The second of these squares approximates to 74, and we seek a small fraction such 13. To divide a given number into three parts such that the sum of any two of the parts gives a square. 6561 841 Therefore the two squares into which 10 is divided are these lies in fact between 2 and 24. II. We have next to divide the square into two squares, one of which, which The three required squares into which to is divided are therefore 1849 2624400 1893401 841 707181 707281 And if we subtract 1 from the first, 3 from the second and 4 from the third, we obtain as the required parts of unity 140447 501557 64377 707181' 707181' 707181 Since the sum of each pair of parts is a square less than 10, while the sum of the three pairs is twice the sum of the three parts or 20, we have to divide 20 into three squares each of which is < 10. But 20 is the sum of two squares, 16 and 4; and, if we put 4 for one of the required squares, we have to divide 16 into two squares, each of which is < 10, or, in other words, into two squares, one of which lies between 6 and 10. This we learnt how to do' [v. 10]. We have, when this is done, three squares such that each is < 10, while their sum is 20; and by subtracting each of these squares from 10 we obtain the parts of 10 required. 14. To divide a given number into four parts such that the sum of any three gives a square. Given number 10. = Three times the sum of the parts the sum of four squares. which is < 10. (1) If we use the method of approximation (rapiσÓτNS), we have to make each square approximate to 7; 1 Wertheim gives a solution in full, thus. Let the squares be x2, 16-x2, of which one, x, lies between 6 and to. 841 6400 7056 841' 841 The required three squares making up 20 are 4, Subtracting these respectively from 10, we have the required parts of the given number |