178 HISTORY OF THE THEORY OF NUMBERS. [CHAP, iv 



interpret as the hypotenuse the letter which stood for one leg, we have a 

 new problem solved by A. Cunningham. 86 



Fermat 87 proposed that St. Martin find two right triangles whose areas 

 are in a given ratio and such that the two legs of the larger triangle differ 

 by unity. 



Fermat 8 ' 8 noted that if in (205769, 190281, 78320) we add the area to 

 the square of the sum of the legs, we get a square. 



Frenicle 89 stated the last result without comment; also that the sum 

 of the area and hypotenuse of (17, 144, 145) is a square; while the first 

 three right triangles in which the sum of the area and smaller leg is a square 

 are (3, 4, 5), (16, 30, 34), (105, 208, 233). 



J. de Billy 90 treated a large number of problems on rational right tri- 

 angles. In the first 44, a prescribed multiple of the area when added to 

 or subtracted from certain sides gives squares. The next five involve 

 the perimeter. In Prob. 58, the cube of the sum of the hypotenuse and 

 one leg when increased by a given multiple of the area shall be a cube, 

 while 55-67 are analogous. In Prob. 68, the areas of (30-2 3n , 18-2 3n , 

 24-2 3n ) are seen to form a geometrical progression of ratio 2 6 , while 69-73 

 are similar. The 120 problems of Ch. 2 do not involve areas, but make 

 certain functions of the sides squares and cubes. 



J. Ozanam 91 found that in the right triangle whose sides are the ratios 

 of 2264592, 18325825 and 18465217 to 20590417 each side exceeds double 

 the area by a square. This problem was proposed in obscure verse in the 

 Ladies' Diary for 1728 as Question 133; a modified uninteresting problem 

 was solved in 1729. 



C. Wildbore 92 took x and 1 x as the legs; they exceed the double area 

 by x z and (1 a;) 2 . Equating the hypotenuse h to v(l x) + x, we get 

 x = (1 - t>2)/(l + 2v - v 2 ). The condition h - (x - x 2 ) = D becomes 

 1 + 4v 3 y 4 = D. First, take v = b/a, b = d 3, a = d + 5; then 

 4d 4 +== D = (2d? - 260d - 2) 2 for d = 4223/66, which yields Oza- 

 nam's answer. The next value of v is said to be 491050/555466, which 

 gives x = 8426546832/76616941657. Elsewhere 93 he took 



1 + 4w 3 - ^ = (1 + raw 2 ) 2 . 

 By the radical in the solution for v, 



2(1 - - n)(2 + n + n 2 ) = D = 4r 2 (l - - ri)\ 

 say. Solving for n, we see that 4r 4 + 12r 2 7 = D. Take r = a/b, 



86 Math. Quest, and Solutions, 3, 1917, 79-80. 



87 Oeuvres, II, 252, letter to Mersenne, Feb. 16, 1643. 



88 Oeuvres, II, 263 (260, 3), letter to Mersenne, Sept. 1, 1643. 



89 Methode pour trouver la solution des problemes par les exclusions, Ouvrages de Math., 



Paris, 1693; M6m. Acad. R. Sc. Paris, 5, 1666-99 (1676), 6d. 1729, 56. 



90 Diophanti Redivivi, Lvgdvni, 1670, Pars Prior, pp. 1-302. 



91 Nouveaux e'le'mens d'algebre, 1702, 604. 



82 Ladies' Diary, 1772, 40-1, Quest. 638; T. Leybourn's Math. Quest, from Ladies' Diary, 

 2, 1817, 342-5; C. Button's Diarian Miscellany, 3, 1775, 356-7. 



93 C. Button's Miscellanea Math., London, 1775, 163-4; Leybourn's Math. Quest. L. D., 

 2, 1817, 342-5. 



