CHAP. IV] PROBLEMS ON THE SIDES OF RIGHT TRIANGLES. 185 



by hypothesis, CE = 8C = 2be 2 + 2bae. Hence 



BE 2 = BC 2 + CE 2 = b 2 f, / = a 4 + 5e 4 + 6aV + 8ae 3 . 



It remains to make / a square, which Wallis suspected to be impossible. 

 Frenicle (Dec. 20, 1661) took a = 2, e = 4, whence / = 52 2 [whereas we 

 desire a > e~], Fermat 139 formed the first triangle from N + 1 and 2. 

 Then the legs of the second triangle are N 2 + 2N + 5 and 4N +12; by 

 their sum of squares, 



(ro \2 



13 + ^N -N 2 ), 



say. Thus N = 1525/546. Hence we use as the first triangle that 

 formed from -f- 979 and 2 546. The resulting triangles are 



(2150905, 2138136, 234023), (2165017, 2150905, 246792). 



If we had used the sum of the legs instead of their difference, we would 

 obtain the simpler solution (1517, 1508, 165) and (1525, 1517, 156). 

 T. Pepin 140 noted that the initial problem is equivalent to 



(3) x 2 + y z = z 2 , u 2 + v 2 = x 2 , u-v = x-y>0. 



We have u, v = a 2 e 2 , 2ae-, x = a 2 + e 2 . According as u is odd or even, 

 y = 2e(a + e) or 2a(a e). Then the first condition becomes 



according as the larger leg of the smaller triangle is odd or even. Contrary 

 to Frenicle's solution a = 2, e = 4, the geometry requires a > e. But 

 we can satisfy the first condition by taking x = d(m 2 n 2 ), y = 2dmn, 

 z = d(m 2 + n 2 ), where d = 1 if x, y are relatively prime, and d = 2 if 

 x, y are even, while m, n are relatively prime and one is even. Then 

 2dmn = 2e(a + e), which is completely solved by 



(4) m = a/3, n = hk, e = (3k, a + e = ah, or e = ah, a + e = 2@k, 



according as d 1 or d = 2, where a, /?, h, k are relatively prime in pairs, 

 the first three being odd and k even. Whether d = 1 or d = 2, 



d(m 2 n 2 ) = a 2 + e 2 

 gives 



(5) k 2 (h 2 + 2/3 2 ) - 2a(3hk + a 2 (h 2 - /3 2 ) = 0. 



Solving this for k/a or hf(3, and making the radicals rational, we get 

 2jS 4 h 4 = D, a 4 2k 4 = D, which have been completely solved by 

 Lagrange 54 of Ch. XXII, so that we know all solutions under a given limit. 

 Then (4) give solutions of the proposed problem. We may also solve (5) 

 by a method equivalent to that of Euler 143 ~ 145 of Ch. XXII. Set h/p = , 

 kja = r?; then 



- 1 = 0. 



139 Inventum Novum of de Billy, in S. Fermat's Diophanti Alex. Arith., 1670, 34-35. Oeuvres 



de Fermat, 3, 1896, 393-4; 4, 1912, 132. 



140 Atti Accad. Pont. Nuovi Lincei, 33, 1879-80, 284-9; extract in Oeuvres de Fermat, 4, 



1912, 219-220. 



