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



Call 77, t] the values corresponding to the same ; and , ' the values 

 corresponding to the same rj. Hence 



Hence all solutions follow from the primitive solution rj = 0, = 1 : 



2 JL -i 1343 



lj * " 3 ; 13' ^ : 113 ; ' 1525' 



The second set is said to furnish the least positive solution of (3) : 



x = 2150905, y = 246792, z = 2165017, u = 2138136, v = 234023. 



M. Martone 141 satisfied the first equation (3) by taking x = 2ab, 

 y = a 2 b 2 , z 2 = a 2 + 6 2 . From the square of the third given equation, 

 we get x 2 2uv = z 2 2xy. Thus we have uv and u v expressed in 

 terms of a, b. Thus 



2v = a 2 - - 2ab - - 6 2 r, r z = 8a 2 6 2 - (2ab - a 2 + 6 2 ) 2 . 



Taking a = 56, we get r 2 = 46 4 , (y, M) = (66 2 , - 8& 2 ) or (86 2 , - 66 2 ). 



MISCELLANEOUS PROBLEMS INVOLVING THE SIDES, BUT NOT THE AREA. 



Diophantus V, 25 relates to xlxfal x] = D for i = 1, 2, 3. A solution 

 will be Xi = tbi/hi if three right triangles (pi, b i} hi) are found such that 

 h-jijiz = t~bib 2 b z . He took (3, 4, 5) as the first triangle and b 3 = 4. From 

 the triangles (13, 5, 12) and (5, 3, 4), the ratio of whose areas is 5 : 1, we 

 can find two triangles such that the product of the hypotenuse and base of 

 one is 5 times that of the other. Indeed, 142 he knew how to deduce from 

 a right triangle (a, ft, 7) a triangle (a, &, c) with ac = 187/2, where a and a 

 are the hypotenuses. He took a = a/2, b = (/3 2 7 2 )/(2a), c = fly/a. 

 From (13, 5, 12) and (5, 3, 4) he thus deduced (6, 119/26, 60/13) and 

 (2|, 7/10, 12/5), the product of the hypotenuse and final leg being 30 and 

 6, respectively. Fermat 143 gave two such triangles for which the ratio in 

 question is 5 : 1, the sides being numbers of 10 and 11 figures (Oeuvres, I, 

 325; 111,263). 



Fermat, 144 to find two right triangles (p, b, h), (p r , b', h') for which 

 p b = b f In! and b h = p' b', took three squares r 2 , s 2 , t z in arith- 

 metical progression and formed the triangles from r + s, s and s + t, s. 

 From r = 1, s = 5, t = 7, we get (11, 60, 61), (119, 120, 169). We may 

 also take r = 7, s = 13, t = 17. 



141 Sopra un problema di analisi indeterminata, Catanzaro, 1887. 



142 Restoration of the obscure text by J. O. L. Schulz, "Diophantus," 1822, 546-61. 



143 P. Tannery, Bull. Math. Soc. France, 14, 1885-6, 41-5 (reproduced in Sphinx-Oedipe, 4, 



1909, 185-7), concludes that Fermat was aided by chance in obtaining his solution, 

 which is not general and contains an error of sign. S. Roberts, Assoc. frang. av. sc., 

 15, II, 1886, 43-9, discussed the problem. Both papers are reprinted in Oeuvres de 

 Fermat, 4, 1912, 168-180. This problem of Format's has been treated by A. Holm 

 and A. Cunningham, Math. Quest. Educ. Times, (2), 11, 1907, 27-29; special cases 

 by K. J. Sanjana and Cunningham, ibid., (2), 13, 1908, 24-26; E. Fauquembergue, 

 1'interme'diaire des math., 24, 1917, 30-1; cf. 25, 1918, 130-1. 



144 Oeuvres, II, 225, letter to Frenicle, June 15, 1641. Cf. II, 229, 232. 



