But 3 should have to 5 the ratio of a square to a square. number such that the difference between the square Form a right-angled triangle from (m, ; The auxiliary triangle must therefore be formed from 24, , and the auxiliary number sought is 4. Put now for the original triangle (hx, px, bx), where (h, p, b) is the right-angled triangle formed from 44,4; this gives pbx2+5=13888o x2, and we have the solution. 3600 [The perpendicular sides of the right-angled triangle are 4. To find a right-angled triangle such that its area minus a given number makes a square. Given number 6, triangle (3x, 4x, 5x), say. Therefore 6x2 – 6 = square = 413, say. Thus, in this case, we must find a right-angled triangle and a number such that = (area of triangle) - (number) of a square. Form a triangle from m, I m = m Therefore m§, and the auxiliary triangle is formed from (,), the auxiliary number being 1. We start again, substituting for 3, 4, 5 in the original [The auxiliary triangle is (4015, 2, 4177), whence so that the required triangle is (4015, 16, 4177).] 5. To find a right-angled triangle such that, if its area be subtracted from a given number, the remainder is a square. Given number 10, triangle (3x, 4x, 5x), say. Thus 10-62= a square; and we have to find a rightangled triangle and a number such that (area of triangle) + (number)2 = of a square. I Form a triangle from m, the area being m2 m or again 65m2 +25= a square = (8m+5)2, say, The rest is obvious. m = 80. 6. To find a right-angled triangle such that the area added to one of the perpendiculars makes a given number. Given number 7, triangle (3x, 4x, 5x). Therefore 6x2+3x=7. In order that this might be solved, it would be necessary that (half coefficient of x)2+ product of coefficient of x2 and absolute term should be a square; but (1)2+6.7 is not a square. Hence we must find, to replace (3, 4, 5), a right-angled triangle such that (one perpendicular)2 +7 times area a square. Let one perpendicular be m, the other I. 7. Therefore 3m + 1 = a square, or 14m + 1 = a square.) and putting, as usual, 7a = 14m + 1, we have m=4. The auxiliary triangle is therefore (24, 1, 2) or (24, 7, 25). and x=1. We have then (6, 7, 25) as the solution'. To find a right-angled triangle such that its area minus one of the perpendiculars is a given number. Given number 7. As before, we have to find a right-angled triangle such that (one perpendicular) + 7 times area a square; this triangle is (7, 24, 25). Let then the triangle of the problem be (7x, 24x, 25x). x=}, and the problem is solved?. 1 Fermat observes that this problem and the next can be solved by another method. "Form in this case," he says, "a triangle from the given number and 1, and divide the sides by the sum of the given number and 1; the quotients will give the required triangle." In fact, if we take as the sides of the required triangle The solution is really the same as that of Diophantus. * Similarly in this case we may, with Fermat, form the triangle from the given number and 1, and divide the sides by the difference between the given number and 1, and we shall have the required triangle. In VI. 6, 7, Diophantus has found triangles §, §, ʼn (§ being the hypotenuse), such that (1) ¦ En+E=a, 8. To find a right-angled triangle such that the area added to the sum of the perpendiculars makes a given number. Given number 6. Again I have to find a right-angled triangle such that (sum of perpendiculars)2 + 6 times area = a square. Let m, I be the perpendicular sides of this triangle; therefore (m + 1)2 + 3m = {m2 + 34m + 4 = a square, while m2 + 1 must also be a square. Therefore m2 + 14m+1) 1} are both squares. m2 + 1) and the auxiliary triangle is (8, 1, §), or (45, 28, 53). x is rational [=], and the solution follows. 9. To find a right-angled triangle such that the area minus the sum of the perpendiculars is a given number. Given number 6. As before, we find a subsidiary right-angled triangle such that (sum of perpendiculars)2+6 times area a square. This is found to be (28, 45, 53) as before. Taking (28x, 45x, 53x) for the required triangle, x=, and the problem is solved'. IO. To find a right-angled triangle such that the sum of its area, the hypotenuse, and one of the perpendiculars is a given number. observing that Diophantus and Bachet appear not to have known the solution, but that it can be solved "by our method." He does not actually give the solution; but we may compare his solutions of similar problems in the Inventum Novum, e.g. those given in the notes to VI. 11 and VI. 15 below and in the Supplement. The essence of the method is that, if the first value of x found in the ordinary course is such as to give a negative value for one of the sides, we can derive from it a fresh value which will make all the sides positive. 1 Here likewise, Diophantus having solved the problem — En - ( + n) = a, Fermat enunciates, as to be solved by his method, the corresponding problem + n - 15 Given number 4. If we assumed as the triangle (hx, px, bx), we should have pbx2+hx + bx = 4 ; and, in order that the solution may be rational, we must find a right-angled triangle such that (hyp. + one perp.)2+4 times area = a square. Form a right-angled triangle from I, m + 1. Then (hyp. + one perp.)2 = 4 (m2 + 2m + 2 + m2 + 2m)3 = m2 + 4m3 + 6m2 + 4m + 1, and 4 times area = 4 (m + 1) (m2 + 2m) Therefore = · 4m3 + 12m2 +8m. m* + 8m3 + 18m2 + 12m + 1 = a square = (6m +1 − m2)3, say, whence m, and the auxiliary triangle is formed from (1, g) or (5, 9). This triangle is (56, 90, 106) or (28, 45, 53). We assume therefore 28x, 45x, 53x for the original triangle, and we have 630x+81x = 4. Therefore x=18, and the problem is solved. II. To find a right-angled triangle such that its area minus the sum of the hypotenuse and one of the perpendiculars is a given number. Given number 4. We have then to find an auxiliary triangle with the same property as in the last problem; therefore (28, 45, 53) will serve the purpose. We put for the triangle of the problem (28x, 45x, 53x), and we have 630x2 - 81x = 4; r, and the problem is solved1. 1 Diophantus has in VI. 10, 11 shown us how to find a rational right-angled triangle S, E, n ( being the hypotenuse) such that Fermat, in the Inventum Novum, Part 111. paragraph 33 (Oeuvres de Fermat, III. p. 389), propounds and solves the corresponding problem In the particular case taken by Fermat a=4. He proceeds thus: First find a rational right-angled triangle in which (since a= 4) |