CHAP, xiv] THREE SQUARES IN ARITHMETICAL PROGRESSION. 437 



by 2 give squares, the sum of the first and third of which is also a square. 

 Take q = l in Frenicle's set; we get p 2 +2p 1, p 2 +l, p 2 2p 1. Hence 

 the conditions are p 2 +3 = D , 2p 2 +2 = D . Set p = m+l, and let the second 

 equal (nm + 2) 2 , whence m = 4 ( 1 ?i) / (n~ 2) . Then 



if n = 5/4, whence p = 23/7. 



To find three squares in A. P. such that any root plus unity is a square, 

 H. Clay 17 took a; 2 , a 2 x 2 , b 2 x 2 as the squares. Set #+l = (r+l) 2 . Then 

 ax+l = (sr+l) 2 determines r. Then &z+l = D if a certain quartic in s 

 is a square, which is the case if s = (2pq 4b)/(q 2 1). Finally, choose a 

 and b so that 1, a 2 , b 2 are in A. P. A. B. Evans 18 took a = 5, 6 = 7 and 

 proceeded similarly. S. Bills 19 employed the numbers a, b = 2pq(p 2 q 2 ); 

 c = p 2 +q 2 , whose squares are in A. P., and took ax/y 2 , bxfy 2 , cxfy 2 as the 

 roots of the required three squares. Then ax+y 2 = (r+y} 2 , ~bx-\-y 2 = (s+y) 2 

 determine x, y. Then cx-\-y 2 = D if a quartic in r is a square, which is the 

 case if r = s(a+b c)/(26). W. J. Miller 19 called the numbers x, y, z and 

 set x+l=m 2 , y + l=n 2 , z + 1 =p 2 , m+n = r(n+p), m n = s(n p}, whence 



m n _ p _1 



rs+2rs r-fs 2 r+s k' 



Then x 2 +z 2 = 2y 2 reduces to k 2 =f(r, s), which is solved. D. T. Griffiths 20 

 took x 2 !, y 2 !, z 2 ! as the numbers. Their squares are in A. P. if 

 x 2 +y 2 -2 = a(y 2 +z 2 -2), y 2 -z 2 = a(x 2 -y 2 }. Taking a = 1/2 (the value when 

 the squares are 1, 5 2 , 7 2 ), and eliminating z, we get 5x 2 y 2 = 4:. This holds 

 if x = 5, y ll, whence 2 = 13. 



To find three squares in A. P. such that each less its root is a square, 

 Smyth 21 took a 2 .x 2 , 6 2 2 , c 2 o: 2 , p l/a, etc. Then x 2 px, are made 

 squares in the usual way. " Epsilon " used the numbers l/X, a/X, b/X, 

 where X = 2x-x 2 and where 1, a 2 , b 2 are in A. P. Now 1/X 2 -1/X=D. 

 Again, t 2 -t=D if t=(k+l) 2 /(4:kl). It is shown that a/X and b/X are of 

 the latter form if 



1 = {4:ab-(ab-a-b) 2 } 2 



X~8ab(ab+b-a)(ab+a-b)(a+b-abY 



To find three squares in harmonical progression the sum of whose roots 

 is a given biquadrate d 4 , "Epsilon" 22 took a, c = 2mn(m 2 n 2 ) and 

 b = m 2 +n 2 . Then the squares of h/a, h/b, h/c are in harmonical progression; 

 equating their sum to d 4 , we get h. 



A. Guibert 23 noted that the common difference of 3 squares in A. P. 

 is a multiple of 24, and similar theorems. The general solution in positive 

 relatively prime integers of a 2 +c 2 = 26 2 is stated to be 



17 The Gentleman's Math. Companion, London, 5, No. 25, 1822, 151-4. 



18 Math. Quest. Educ. Times, 16, 1872, 27-28. 



19 Ibid., 11, 1869, 88-91. 



20 Ibid., 63, 1895, 46-7. 



21 The Gentleman's Math. Companion, London, 5, No. 26, 1823, 214-8. 



22 The Gentleman's Math. Companion, London, 5, No. 28, 1825, 365-6. 



23 Nouv. Ann. Math., (2), 1, 1862, 213-9. 



