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



From one right triangle (x, x + 1, z) whose legs are consecutive integers, 

 Fermat 110 deduced the second triangle (X, X + 1, Z), where 



X = 2z + 3x + 1, Z = 3z + 4x + 2. 



For example, we have the series (3, 4, 5), (20, 21, 29), (119, 120, 169), 

 The alternate triangles give solutions of the problem to find right triangles 

 whose least side differs from the other two sides by squares. He noted 

 later (pp. 232-3) that such a triangle is formed from r 2 + s 2 , 2s(r s). 



Fermat 111 noted that the sixth such triangle is (23660, 23661, 33461). 

 From the first such triangle (3, 4, 5), we get the second by taking the double 

 (viz., 24) of the sum of the three sides and subtracting separately the legs 

 and adding the hypotenuse. 



J. Ozanam 112 gave the first six such triangles. If one is formed (Dio- 

 phantus 7 ) from m, n, where m > n, the next is formed from m, 2m + n. 

 In the edition by J. E. Montucla, 1, 1790, 48, the triangle is formed from 

 any two consecutive terms of 1, 2, 5, 12, 29, 70, -, k, where k is such that 

 one of the two numbers 2/c 2 1 is a square. The same rule was given 

 by Griison. 22 



C. Hutton 113 noted that, if p r /q r is the rth convergent to V2, then 

 p r pr+i and 2q r q r +i are consecutive integers the sum of whose squares is a 

 square ql r+z . 



Du Hays 114 gave triangles the difference of whose legs is 1 (or 7). 



L. Brown 115 gave the first six and the eleventh such triangles. 



G. H. Hopkins and M. Jenkins 116 reduced the problem to z 2 2y* = 1, 

 and gave recursion formulas for the solutions. A. B. Evans used the con- 

 tinued fraction for V2. Cf. Moret-Blanc 154 of Ch. XII. 



Judge Scott 117 gave the first eight and the eleventh. 



A. Martin 118 employed the legs %(x 1), whence x 2 2y 2 = 1, and 

 the odd convergents x n /y n to the continued fraction for V2". Thus 

 x n = 6z n -i x n -2 and likewise for the y's. Also 



2x n , 



The eightieth such triangle is found. 



T. T. Wilkinson stated and J. Wolstenholme 119 proved a rule equivalent 

 to a recursion formula for the solutions of x 2 2y 2 = 1. 



D. S. Hart 120 took x and x + 1 as the legs. Then 



2x z + 2x + 1 = D = (xpfq - I) 2 



110 Oeuvres, II, 224-5. Reproduced in Sphinx-Oedipe, 7, 1912, 103-4. 



111 Oeuvres, II, 258; letter to'St. Martin, May 31, 1643; reproduced, Sphinx-Oedipe, 7, 1912, 



104. 



112 Recreations Math., 1, 1723; 1724; 1735, 51; etc. (first ed., 1696). 



113 English transl. of Ozanam's Recreations, 1, 1814, 46. 



114 Jour, de Math., 7, 1842, 331. 



115 Math. Monthly (ed., Runkle), Cambridge, Mass., 2, 1860, 394. 

 16 Math. Quest. Educ. Times, 12, 1869, 104-6. 



117 Of commensurable right-angled triangles . . . , Bucyrus, Ohio, 1871, 23 pp. 



118 Math. Quest. Educ. Times, 14, 1871,89-91; 16,1872,107; 19,1873,89; 20,1874,21,42-4. 



119 Ibid., 20, 1874, 97-99. 

 20 Ibid., 63-4. 



