622 HISTORY OF THE THEORY OF NUMBERS. [CHAP, xxn 



Euler 49 noted that z 4 -2?/ = (p 2 -2g 2 ) 2 for y* = 2pq, z 2 = p 2 +2# 2 . The 

 latter holds if p = r--2s-, q = 2rs. Then 2pq = y*= 4rs(r 2 -2s 2 ). Set 

 r = t z , s = u 2 . For the upper sign, t 4 2u*= D, whereas t and u are smaller 

 than x, y. Hence take the lower sign. Thus a solution of 2u* 4 =D 

 yields a solution of z 4 2?/ 4 =n. For t = u = l, we get re = 3, y = 2. Then 

 for t = 3, u = 2, we get x = 113, ?/ = 84. Again, it = 13, = 1 gives x = 57123, 

 7/ = 6214. Lebesgue 56 (end) noted that this solution is incomplete. 



Euler 50 treated x+y=H, z 2 +?/ 2 =0 2 +l) 4 by taking .T = z 4 -6z 2 +l, 

 ?/ = 42 3 40. Then x+y is the product of the two factors 2 2 +(22 V2)z 1, 

 which he equated to (z-\-pq V2) 2 . By the rational and the irrational parts, 

 we get 



pq 0dbV2o 4 -l 



? l * 71 -L 



* 1-q' P ~ 1+q ' 

 Thus q = 13 gives p = 18 or -113/7, q=- 13 gives p = 21 or 113/6. 



Euler 51 reduced (2) to (7) by setting v = 2x*+y*, whence z 4 +8(xy) 4 = v-. 

 Conversely, let <? 4 +8p 4 = r 2 ; then 8p 4 =(r+q 2 )(r g 2 ), so that q and r are 

 odd. First, let r+g 2 = 2o:, r <? 2 = 4/3, where a is odd. Then p 4 = a/3, and 

 a, 18 are relatively prime; whence a = s 4 , j3 = Z 4 , p = sZ. By subtraction, and 

 cancellation of 2, g 2 = s 4 2 4 . Second, let rq- = 2a, r-|-<? 2 = 4/3, where is 

 odd. Proceeding as before, we get q' 2 = 2t 4 s 4 . While in the second case 

 only we obtained (2), the reduction can always be made since / 4 +S0 4 = /i 2 

 implies 2x 4 y* = z 2 for 



x=/ 2 +2fg*-gh, y=f*-4fg*+gh, z=/ 6 +/V-6/^+24/y-8/. 

 In quoting this solution, Lebesgue, 56 p. 74, gave f 2 g incorrectly for fg- in x. 



Euler 52 noted that x+y = B 2 , x 2 +y 2 = A* imply (x-y)* = 2A 4 -B 4 . The 

 latter is the square of r ? 2 +2^-^ 2 if A 2 = 2 +77 2 , 3"- = (^^-2^. Taking 

 ij = 2a6cd, we have A = a 2 6 2 +c 2 d 2 if ^ = a 2 6 2 -c 2 ^ 2 , and 5 = a 2 c 2 -26 2 d 2 if 

 ^+?7 = a 2 c 2 +26 2 d 2 . The two values of +77 are equal if 



a _bcr d_ 6c=Fr 2_ow 4 



3 = ~9 t9 ^ r ~ 7o , ^ T~ = 2u C. 



a c 2 o 2 a 2o 2 +c 2 



Hence 2A 4 J5 4 is a rational square if 26 4 c 4 is. Taking 6 = c = l, we have 

 a = 3 ? d = 2, = 5, 77 = 12, A = 13, B = l, 2-13 4 -l = 239 2 ; since B<A, x 

 and T/ are n t both positive. Taking 6 = 13, c = l, we have r = 239, 

 afd= -3/2 or 113/84. For a = 3, d= -2, then = 1517, r? = - 156, A = 1525, 

 J5= 1343, which 53 do not yield positive x, y. For a = 113, d = 84, then 

 A = 2165017, B= 2372159, and we obtain very large solutions. [In fact, 

 Fermat's (1). Since x y = r? 2 +2?7 2 , x-\-y = B 2 , we have x = 2^rj ) y = - 7? 2 . 

 Thus x, y are the legs of the right triangle formed from , 77. Here 

 = 2150905, 77 = 246792, as in Fermat's solution.] 



49 Algebra, II, art. 211; French transl., pp. 260-3; Opera Omnia, (1), I, 444-5. 



60 Opera postuma, 1, 1862, 491 (about 1774). 



61 Opera postuma, 1, 1862, 221-2 (about 17SO). 

 "Opusc. anal., 1, 1783 (1773), 329; Comm. Arith., II, 47. 



"The method of Euler, Algebra, 2, art. 140, to make 2.c 4 1 a square does not give all solu- 

 tions since 1525/1343 is omitted (remarked by Lebesgue 66 ). E. Fauquembergue, 

 Pinterm6diaire des math., 5, 1898, 94, claimed to prove that x = l, x = 13 are the only 

 integral solutions. 



