25. To find three squares such that the product of any two minus I gives a square. This reduces, similarly, to v. 22 above. 26. To find three squares such that, if we subtract the product of any two of them from unity, the result is a square. This again reduces to an earlier problem, v. 23. 27. Given a number, to find three squares such that the sum of any two added to the given number makes a square. Given number 15. Let one of the required squares be 9; I have then to find two other squares such that each Let one pair of factors be 4/x, 6x, and let the side of one square be half their difference or - 3x. 2 x Let the other pair of factors be 3/x, 8x, and let the side of the other square be half their difference or Therefore each of the squares + 24 gives a square. It remains that their sum +15 = a square; therefore 2 + 1 +15= a square, Equating the square of half the sum of the factors to the larger expression, we have But even here, as the value of x which we have found is negative, we ought, strictly speaking, to deduce a further value by substituting y 1218311 74649600 for x in the equations and solving again, which would of course lead to very large numbers. 1 The text adds the words "and [let us take] sides about the right angle in a rightangled triangle." I think these words must be a careless interpolation: they are not wanted and give no sense; nor do they occur in the corresponding place in the next problem. or = 61 + +25x3-9= a square = 25×3, say. Therefore x, and the problem is solved1. 28. Given a number, to find three squares such that the sum of any two minus the given number makes a square. Given number 13. Let one of the squares be 25; I have then to find two other squares such that each +12= a square, and (sum of both) - 13 = a square. Divide 12 into factors in two ways, and let the factors be (3x, 4/x) and (4x, 3/x). Take as the sides of the squares half the differences of the factors, i.e. let the squares be (14x-2)', (2x − 1)". Each of these + 12 gives a square. It remains that the sum of the squares - 13 = a square, 1 Diophantus has found values of §, n, † satisfying the equations n2+52+a= u2) 2+ y2+a=w2) Fermat shows how to find four numbers (not squares) satisfying the corresponding conditions, namely that the sum of any two added to a shall give a square. Suppose a = 15. 1 529 Take three numbers satisfying the conditions of Diophantus' problem, say 9. 100 225 Assume x2 - 15 as the first of the four required numbers; and let the second be 6x+9 (because 9 is one of the square numbers taken and 6 is twice its side); for the same 46 529 reason let the third number be and the fourth x+ 15 225 5 I 100 Three of the conditions are now fulfilled since each of the last three numbers added to the first (x2 - 15) plus 15 gives a square. The three remaining conditions give the tripleequation (+9)* * Fermat observes that four numbers (not squares) with the property indicated can be found by the same procedure as that shown in the note to the preceding problem. If a is the given number, put r2+a for the first of the four required numbers. 29. To find three squares such that the sum of their a square. Let the squares be x2, 4, 9 respectively1. Therefore +97 a square (x2 - 10)2, say; = = squares is If the ratio of 3 to 20 were the ratio of a square to a square, the problem would be solved; but it is not. Therefore I have to find two squares (p2, qa, say) and a number (m, say) such that m2 -p' — q' has to 2m the ratio of a square to a square. Let p2, q2=4 and m=+4. 2 Therefore m2-p1 − q♦ = (≈2 + 4)2 — 2′ — 16 = 8%a2. Hence 882/(28+8), or 422/(≈2+4), must be the ratio of a square to a square. therefore z = 1, and the squares are p2= 2, q2 = 4, while m=61; or, if we take 4 times each, p2 = 9, q2 = 16, m = 25. Starting again, we put for the squares x2, 9, 16; then the sum of the squares = x + 337 = (x2 - 25), and x = 12. The required squares are 144, 9, 16. 25 30. [The enunciation of this problem is in the form of an epigram, the meaning of which is as follows.] A man buys a certain number of measures (xóes) of wine, some at 8 drachmas, some at 5 drachmas each. He pays for them a square number of drachmas; and if we add 60 to this number, the result is a square, the side of which is equal to the whole number of measures. Find how many he bought at each price. Let x = the whole number of measures; therefore 2 – 60 was the price paid, which is a square = (x-m), say. If now k2, 12, m2 represent three numbers satisfying the conditions of the present problem of Diophantus, put for the second of the required numbers 2ka +k2, for the third 2/x+12, and for the fourth 2mx+m2. These satisfy three conditions, since each of the last three numbers added to the first (x2+a) less the number a gives a square. The remaining three conditions give a triple-equation. "Why," says Fermat, "does not Diophantus seek two fourth powers such that their sum is a square? This problem is in fact impossible, as by my method I am in a position to prove with all rigour." It is probable that Diophantus knew the fact without being able to prove it generally. That neither the sum nor the difference of two fourth powers can be a square was proved by Euler (Commentationes arithmeticae, 1. pp. 24 sqq., and Algebra, Part II. c. XIII.). H. D. Now of the price of the five-drachma measures+of the price of the eight-drachma measures = x; so that x3-60, the total price, has to be divided into two parts such that of one + of the other = x. We cannot have a real solution of this unless or x" < 8x+60 x2= 8x + some number less than 60, whence x is not greater than 12. Now (from above) x = (m2 + 60)/2m; therefore 22m < m2 + 60 < 24m. Thus (1) 22m = m2 + (some number less than 60), (2)_24m = m2 + (some number greater than 60), and therefore m is less than 21. Hence we put m = 20, and x2- 60= (x-20), so that x = 114, x2 = 1324, and x2 – 60 = 72}. Thus we have to divide 72 into two parts such that of one part plus of the other = 11}. Let the first part be 52. Therefore (second part) = 111 – 2, or second part = 92-8z; therefore 52 +92-82 = 72, BOOK VI To find a (rational) right-angled triangle such that the hypotenuse minus each of the sides gives a cube'. Let the required triangle be formed from x, 3. Therefore hypotenuse = +9, perpendicular = 6x, base = x2-9. Thus +9-(1-9)= 18 should be a cube, but it is not. Now 182.32; therefore we must replace 3 by m, where 2. m2 is a cube; and m = 2. We form, therefore, a right-angled triangle from x, 2, namely (+4, 4, x-4); and one condition is satisfied. The other gives xa − 4x + 4 = a cube; therefore (x-2) is a cube, or x - 2 is a cube Thus = 10, and the triangle is (40, 96, 104). 2. To find a right-angled triangle such that the hypotenuse added to each side gives a cube. Form a triangle, as before, from two numbers; and, as before, one of them must be such that twice its square is a cube, i.e. must be 2. We form a triangle from x, 2, namely +4, 4*, 4 − x2; Thus x + 2 = a cube which must be < 4 and > 2 = 27, say. and the triangle is (135, 5, 377). or, if we multiply by the common denominator, (135, 352, 377). 3. To find a right-angled triangle such that its area added to a given number makes a square. Let 5 be the given number, (3x, 4x, 5x) the required triangle. 1 Diophantus' expressions are è è rŷ væоTELovy, "the (number) in (or representing) the hypotenuse," ò év ékatépą Tŵr ópoŵr, “the (number) in (or representing) each of the perpendicular sides,” ó év rý iußadê, “the (number) in (or representing) the area," etc. It will be convenient to say "the hypotenuse," etc. simply. It will be observed that, as between the numbers representing sides and area, all idea of dimension is ignored. |