CHAP. XIX] SYSTEMS OF EQUATIONS OF DEGREE Two. 529 



W. Wright 153 noted that sx yz = m 2 , syxz = n 2 , szxy = r 2 , s = x+y+z, 

 lead to the problem (x+y} 2 = m 2 +n 2 , (x+z) 2 = m 2 +r 2 , (y+z) 2 = n 2 +r 2 at 

 the beginning of this Chapter. 



S. Jones 154 made a = sx+yz, (3 = sz+yx, y = sy+zx squares, where 

 s = x+y+z, by taking y = ax, a = b 2 , y = c 2 , whence x = (a 2 +b 2 c 2 )/(2a), 

 etc., and /3= D. 



J. R. Young 155 found four numbers whose sum is a square and such 

 that if unity be added to the product of the sum by any one of them there 

 results a square. Let the numbers be zl, xy. It suffices to make 

 4x, 4x 2 4xy+l squares. Take x = 4 and set 65 lQy = m 2 , 65 + 16y = n 2 . 

 Then ?rc 2 +n 2 = 130, which holds if m = 3, n = ll. 



W. Wright and others found 156 four numbers v, x, y, z whose sum is a 

 square n 2 and such that vn 2 +l = D, etc. Equate vn 2 +l, xn*+l, yn 2 +l, 

 zn 2 +l to the squares of 1+8, 1+r, 1+9, 1+p- By addition, s 2 +2s+l=n* 

 if r 2 +q 2 +p 2 +2r+2q+2p = l. The latter is solved for r after taking 

 q = m l, p = lm l. Several solvers used the numbers x 1 , xy. 



To make x 2 +y 2 +S, x 2 +z 2 +S, y 2 +z 2 +S squares, where 



S = 2xy+2xz+2yz, 



W. Wright 157 noted that the functions factor, being a(b+c), 6(a+c), c(a+6), 

 where a = x+y, b = x+z, c = y+z. Take b = na, c = ma, n(m+l)=n 2 2 , 

 m(n+\} = (n% p} 2 . We get m and n. Then m+n = N/D, where N and 

 D are quadratic in . Take N=(p+q) 2 to get |. Then D = D becomes a 

 quartic in q. C. Holt noted that one condition is satisfied if the numbers 

 are 5nm, mkn, 4n. Baines wrote s = x+y+z; thus s 2 z 2 , s 2 y 2 , 

 s z x 2 are to be squares, say of (s+z)[m, (s+y)/n, (s+x)/r. We get x, y, z. 

 To satisfy Sz = s, take r = 3, n= -37/36, m = 25/21. 



To find three numbers double the sum of any two less the third being 

 a square, double the sum of any two of their squares less the square of the 

 third being a square, while the last three squares have the same property, 

 W. Wright 158 used the numbers x, y, x+y. Then all but the first three 

 conditions are satisfied. Take x+y = a 2 , 4x+y = (2ap) 2 . For the result- 

 ing x, a, 4y+x= D if p 4 +54p 2 y+9y 2 = D = (3y v} 2 , which gives y. 



To make 159 2(v+x+y+z} = D =4a 2 , a = 2(x+y+z} 2 -2v 2 = D, etc., 

 note that a = 4a 2 (x+y+z v). Hence take #+?/ +2 y = 46 2 , etc. The con- 

 dition a 2 = 6 2 +c 2 +d 2 +e 2 is satisfied by taking a = e+r and finding e. 



Several 160 discussed the problem to make a+b, b+c, b c, a+2b+c+d 

 and a 2 +bc+bd+cd squares, (a+b)(b c)=6+c, and b 2 c 2 = l. 



153 The Gentleman's Math. Companion, London, 3, No. 17, 1814, 462-4. 



1M Ibid., 3, No. 18, 1815, 317-8. 



166 Algebra, 1816. Amer. ed. by S. Ward, 1832, 331. 



166 The Gentleman's Math. Companion, London, 5, No. 26, 1823, 240-2. 



167 Ibid., 5, No. 29, 1826, 500-2. 

 158 Ibid., 5, No. 30, 1827, 575-6. 

 169 Ibid 558 



160 Math. Miscellany, Flushing, N. Y., 1, 1836, 154-5. 



35 



