CHAP. XV] LINEAR FUNCTIONS MADE SQUARES. 451 



factor from our numbers. Then A, B, C, D are all positive if 

 a condition expressed in terms of / and z and treated at length by Euler. 

 For z = 4, /= 7/2, we drop the factor 1/2 and get a: = 30, x l = 26, y = 1, 2/1 = 15, 

 p = 839, g = 329, r = 901; multiplying the resulting A, - , D by 4, we get 

 the integral solutions 



722, 432 242, 2 814 962, 3 246 482. 

 J. Leslie 36 made 2+1 = D, w+l = D, s+y+l=given D by setting 



P. Cossali 37 made F = hx-\-n 2 and F+/x squares by taking F=(y+n) 2 , 



thus finding y. Next, if (ad be}/ (a c) is a square r 2 , ax +6 and cx+d are 

 made squares. Set cx+d = (y+r} 2 ; for the resulting x, 



ax+b = - (y 2 +2ry}+r 2 =(py r) 2 . 



C 



If (be ad) I c is a square g 2 , set cx+d = y~; then ax+b = y 2 a/c+q- can be 

 made the square of g %. To make (pp. 145-6) H+x=\3, # z=D, 

 according to L. Pisano, we have only to express 2H as a sum of two squares. 



To find three numbers in geometrical progression the difference of 

 any two of which is a square, R. Nicholson 38 took nx, n-x, n 3 x as the 

 numbers. Since the ratios of their differences are 1 : w+1 : n, take n = v 2 , 

 v 2 + 1 = D = (y+s) 2 . For the resulting v, n I is a square. Taking x to be a 

 square, we get an answer. J. Cunliffe took no, 4 , na 2 b 2 , nb 4 as the numbers, 

 where n = a 2 6 2 ; the single condition 2 +fr 2 =D is satisfied if b-r 2 s 2 , 

 a = 2rs. 



To find three numbers in geometrical progression whose sum is a square, 

 several 39 took x 2 , nx 2 , n 2 x 2 , l+n+n 2 = D = (ne-1) 2 . 



To find 40 three numbers the difference of any two being a square, 

 take x y = IQv 2 , xz = 25v 2 , yz = 9v 2 , where v and z are arbitrary; or take 

 5x 2 , x 2 , b 2 +x 2 , where 4:X 2 b 2 =(2xn) 2 gives x; or take (z+1) 2 , 2z+l, 4x, 

 where 2x 1 = D. 



J. Cunliffe 41 made x y, etc., and x+yz, etc., squares. Take 



x-\-y z = a 2 , x-\-z y = b 2 , y-{-zx = c 2 . 

 Equate x - y = | (b 2 - c 2 ) to e 2 , x - z = | (a 2 - c 2 ) to d\ Then 



y-z = (a 2 -b 2 )=d 2 -e 2 

 must be a square, whence d = 2rs(m 2 -\-n 2 ), e = 2rs(2mri). Set 



(a+b)r = 2s(d+e), (a-b)s = r(d-e), 

 which give 



a= (m 2 +n 2 )(r 2 +2s 2 ) -2mn(r 2 -2s 2 }, b = 2mn(r 2 +2s 2 } - (m 2 +n 2 )(r 2 -2s 2 ). 



36 Trans. Roy. Soc. Edinb., 2, 1790, 193, Prob. IV. 



37 Origine, Trasporto in Italia . . . Algebra, 1, 1797, 105-7. 



38 The Gentleman's Diary, or Math. Repository, 1798, No. 58; Davis' ed., 3, 1814, 290. 



39 The Gentleman's Math. Companion, London, 1, No. 2, 1799, 18. 



40 Ibid., 21. 



41 The Gentleman's Diary, or Math. Repository, London, No. 61, 1801, 43, Quest. 806. 



