CHAP. XXII] 2z 4 -i/ 4 =n; SYSTEM x+y = A-, x 1 -\-y- = B^. 621 



The ratio of x+5 to 12 is that of 2150905 to 246792. The triangle formed 

 from these is (1). He noted that the problem is equivalent to that to find 

 two numbers whose sum is a square and sum of squares is a biquadrate. 



Fermat 42 noted that in the right triangle (156, 1517, 1525) the square 

 of the difference of the legs exceeds the double of the square of the least leg 

 by a square. Without giving details he added that this triangle serves to 

 find a right triangle whose hypotenuse is a square and whose least side 

 differs from the other two by squares. 



Frenicle 43 gave details on the last problem. An analysis followed by 

 numerical trials led him to the triangle, formed from 6 = 156 and a = 1517, 

 having the sides 



2ab = 473304, a- - 6 2 = 2276953, a 2 + b- = 2325625 = 1 525 2 . 



The least side differs from the other two by the squares of 1343 and 1361. 

 As remarked by A. Genocchi 44 these results imply that 2x 4 y 4 = D has the 

 solution = 1525, ?/ = 1343 [Lagrange, 54 Euler 55 (third memoir), Lebesgue 56 ]. 



E. Torricelli 45 proposed the problem to find a right triangle with integral 

 sides whose hypotenuse, sum of legs and sum of hypotenuse and larger leg 

 are all squares. E. Lucas 46 stated that this problem was proposed by 

 Fermat and that its solution depends on z 4 2y 4 = z~. In fact, Fermat 4Ca 

 proposed the problem to Torricelli. An attempt to trace its origin has 

 been made by E. Turriere. 466 Cf. *M. Cipolla. 46c 



J. Ozanam 47 treated the problem of Fermat 37 by the method essential^ 

 the same as employed by L. Euler. 48 If the legs are x, y, then x+y is to 

 be a square and x*-{-y~ a biquadrate. In this form the problem was proposed 

 by Leibniz. Euler made x~-\-y 2 a square (p 2 +<? 2 ) 2 by taking x = p i q i ) 

 y = 2pq. Then p 2 +g 2 is a square for p = r 2 s 2 , q = 2rs, whence 



z 2 +2/ 2 =(r 2 +s 2 ) 4 . 

 It remains to make 



x + y = r 4 + 4r 3 s 6r 2 s 2 4rs 3 + s 4 



a square. It will be the square of r 2 2rs+s 2 if r = 3s/2. Taking r = 3, 

 s = 2, we obtain a negative value 119 for x. Setting r = 3s/2+, we get 



which is the square of s 2 + 148s* -4Z 2 if s/t = 84/1343. Taking s = 84, we get 

 r = 1469 and x, y as in (1). 



42 Oeuvres, II, 265-6, letter to Carcavi, 1644. 



43 Methode pour trouver la solution des problemes par lea exclusions, Ouvrages de math., 



Paris, 1693; MSm. Acad. R. Sc. Paris, 5, 1666-99 [1676]; ed. 1729, 81-5. 



44 Atti R. Accad. Sc. Torino, 11, 1876, 811-29. 



45 G. Loria, I'interm&liaire des math., 24, 1917, 97-8. Cf. 25, 1918, 83. 



46 Bull. Bibl. Storia Sc. Mat. Fis., 10, 1877, 289. 



46a Letter from Mersenne to Torricelli, Dec. 25, 1643, Bull. Bibl. Storia Sc. Mat, Fis., 8, 1875, 



411; Oeuvres de Fermat, 4, 1912, 82-3 (cf. p. 88). 

 466 L'enseignement math., 20, 1919, 245-268. 

 46c Atti Accad. Gioenia sc. nat. Catania, (5), 11, 1919, No. 11. 



47 Nouv. elemene algebre, Amsterdam, 2, 1749, 480-1. 



48 Algebra, 2, 1770, art. 240, pp. 503-5; French transl., 2, 1774, p. 336; Opera Omnia, (1), 



I, 483-4. 



