166 HISTORY OF THE THEORY OF NUMBERS. [CHAP. IV 



Diophantus (III, 22, etc.) referred to the right triangle with these sides as 

 that formed from the two numbers m, n. 



Brahmegupta 8 (born 598 A.D.) gave explicitly the solution (1). 



An anonymous Arabic manuscript 9 of 972 stated that in every primitive 

 right triangle (i. e., with relatively prime integral sides), the sides are given 

 by (1). Necessary conditions that (1) give a primitive triangle are that 

 m, n be relatively prime and m + n be odd. The hypotenuse of a primitive 

 right triangle is a sum of two squares and is of the form 12k + 1 or 12k + 5, 

 though not all such numbers are sums of two squares. But 65 2 is a sum 

 of two squares in two ways: 63 2 + 16 2 = 33 2 + 56 2 . To find a triangle 

 with a given hypotenuse h, we need an expeditious method to find two 

 numbers the sum of whose squares equals h. If the last digit d of h is 1, the 

 two squares end in 5 and 6 or in 00 and 1. If d = 3, they end in 4 and 9; 

 if d = 7, in 1 and 6; if d = 5, in 00 and 5, 1 and 4, or 6 and 9; if d = 9, 

 in 00 and 9, or 4 and 5; with similar rules if d is even. 



The Arab Ben Alhocain 10 (tenth cent.) gave a geometrical proof that 

 (1) give the sides of a right triangle, and noted that if the hypotenuse is 

 even, also both legs are even. Rules equivalent to that by Pythagoras are 

 given; also false theorems on triangles formed from several consecutive 

 numbers. 



Alkarkhi 11 (end of tenth cent.) derived the solution 3, 4, 5 of x 2 + y z = z 2 

 by setting y = x + 1, z = 2x 1. 



Bh^scara 12 (born 1114) gave (1) and employed it, as had Brahmegupta, 

 to find the second leg (m?/n n)/2 and hypotenuse, (m 2 /n + n)/2, given 

 one leg m. Given the hypotenuse h, the legs are 12a I = 2hb/(b z + 1) and 

 Ib - h or h - q and bq, where q = 2h/(b z + 1). To find (p. 201) a right 

 triangle whose area equals the hypotenuse take 3x, 4x, 5x as the sides. 



Leonardo Pisano 13 employed the fact that the sum 1 -f 3 + of n 

 consecutive odd numbers is n 2 to find two squares whose sum is a square. 

 First, if one square a 2 is odd, take the other to be 1 + 3 + + (a 2 2) ; 

 their sum 1 + 3 + + a 2 is a square. If one square is even, as 36, add 

 and subtract unity from its half, obtaining the consecutive odd numbers 

 17 and 19; then 1 + 3 + + 15 = 64 and 



64 + 36 = 1 + + 15 + 17 + 19 = 10 2 . 



8 Brahme-sphut'a-sidd'hdnta; Algebra with Arithmetic and Mensuration, from the Sanskrit 



of Brahmegupta and Bhascara, transl. by H. T. Colebrooke, London, 1817, 306-7, 

 363-72. 



9 French transl. by F. Woepcke, Atti Accad. Pont. Nuovi Lincei, 14, 1860-1, 213-227, 241- 



269 (M. Cantor, Geschichte Math., ed. 3, I, 1907, 751-2). 



10 Ibid., 301-24, 343-56. 



11 Extrait du Fakhri, French transl. by F. Woepcke, Paris, 1853, 89. 



12 Colebrooke, 8 pp. 61-63. John Taylor's transl. of Brahme . . . , 8 Bombay, 1816, p. 71. 

 120 Same in Ladies' Diary, 1745, 14, Quest. 254; T. Leybourn's Math. Quest. Ladies' Diary, 



1, 1817, 366-7; C. Button's Diarian Miscellany, 2, 1775, 200. 



"Liber quadratorum L. Pisano, 1225, in Tre Scritti inediti, 1854, 56-66, 70-5; Scritti L. 

 Pisano, 2, 1862, 253-4. Cf. A. Genocchi, Annali Sc. Mat. Fis., 6, 1855, 234-5; P. Vol- 

 picelli, Atti Accad. Pont. Nuovi Lincei, 6, 1852-3, 82-3; P. Cossali, Origine, Trasporto 

 in Italia . . . Algebra, 1, 1797, 97-102, 118-9. 



