498 HISTORY OF THE THEORY OF NUMBERS. [CHAP, xix 



function derived from B by replacing/ by (/+!)/(/ 1), viz., 



(f~-l)x-2fy 



Then x*+B*+C* = S = x*+y* gives z/?/ = 8/(/ 2 -l)/(/ 2 4-l) 2 . Take y equal 

 to the denominator and multiply all the numbers by/ 2 +l. Hence 



^ = 8/(/ 4 -l), = (l-/ 2 )(.f-14/ 2 4-l), C = 2/(3/ 4 -10/ 2 +3). 



For/=2, we get 240, 117, 44. E. B. Escott 5 also gave the last solution. 

 Euler 6 set x 2 = 4mnpq, y = mp nq,z = np mq. Then 



x 2 -\-y- = (mp-\-nq) 2 , x z -\-z 2 =(np -\-rnq') 2 . 

 For y = 2(m 2 n 2 ~)rs, z = (m 2 n 2 )(r 2 s 2 ), we get 

 ?/ 2 +2 2 =(m 2 n 2 ) 2 (r 2 +s 2 ) 2 , p = 2mrs n(r 2 s 2 ), q = 2nrs m(r 2 s 2 }. 



The resulting expression for x 2 /4 is a quartic function of r which is the 

 square of mnr 2 (w 2 +n 2 )rs+wns 2 if r = 4ran, s = m 2 -}-n 2 . Then 



x = 2mn (3m 2 n 2 ) (3n 2 m 2 ) , y = Smn (ra 4 ?i 4 ) , 

 z = (m 2 n 2 ) (m 2 4mn + n 2 } (ra 2 -f 4mn + n 2 ) , 



which, apart from signs, equal the products of n 6 by Euler's 4 values when 

 f=mfn. The simplest solution arises from ra = 2, n = l: .T = 44, ?/ = 240, 

 2 = 117, whence x~+y~, etc., are the squares of 244, 125, 267. 



From the sum of the roots of three squares, the sum of any two of 

 which is a square, subtract the area of a right triangle; the remainder is a 

 square which if decreased by the sides of the triangle yields remainders 

 which are squares in arithmetical progression. L. Blakeley 7 took 44x, 

 117 x, 24Qx as the roots of the required squares, the sum of 44 2 , 117 2 , 240 2 

 by twos being known to be squares; also let 3?/, 4y, 5y be the sides of the 

 right triangle of area Qy z . Then 40Lr 6?/ 2 = D =a 2 . Also, a- 3y = b z , 

 a 2 -4y = c 2 , a 2 - 5y = d\ Take y = r 2 - 2dr. Then 



The product of the last two is a square if d = r/Q. Then d 2 = r 2 /36, 

 c 2 = 25r 2 /36, & 2 = 49r 2 /36 are in A. P. 



P. Barlow 8 noted that the first part of this question is satisfied if the 

 roots of the squares are 5750/48, 4852/44 and z (from J. Bonnycastle's Algebra, 

 p. 148). Next, we need a square which if diminished by each side of a 

 right triangle the remainders are three squares in A. P., whence the sides 

 of the triangle are in A. P., and hence proportional to 3, 4, 5. Let the 

 squares be (a 2 +2a&-6 2 ) 2 , (a 2 +b 2 ) 2 , (V -\-2ab-a 2 )' 1 , with the common dif- 

 ference 5 = 4ab (a 2 - 6 2 ). Thus let 35, 45, 55 be the sides. Then 



= D = 6--4a6-a 22 



if 8a 3 b-86 3 a = 12a 2 6 2 , i. e., (2a-2b)(a+fe) =3a&, which holds if a = 26. 



6 L'intermddiaire des math., 8, 1901, 103-4. 



6 Posth. paper, Comm. Arith., 2, 1849, 650; Opera postuma, 1, 1862, 103-4. 



7 Ladies' Diary, 1805, p. 43, Quest. 1131; Leybourn's Math. Quest. L. D., 4, 1817, 45-6. 



8 The Diary Companion, Supplement to Ladies' Diary, London, 1805, 45-6. 



