448 HISTORY OF THE THEORY OF NUMBERS. [CHAP, xv 



quartic in d which is the square of l+2d 5d 2 for d = 12/13. C. Wildbore's 

 solution is the same as Landen's second with f=afb, g = x/y. C. Hutton 

 took 4z, 4+ a; 2 , l+4a; 2 as the numbers. Then 5x 2 +5 and 3z 2 3 are to be 

 squares. The product 15z 4 15 is a square for x = 2, and for x=z 2 

 becomes a quartic in z which is made a square by the usual method. He 

 obtained Bumpkin's 22 answer. T. Leybourn 24 took x-\-y = u 2 , x+z = v 2 , 

 y+z = w 2 ; it remains to make u 1 v 2 , v 2 w 2 , u 2 w 2 squares, which is known 25 

 (Lowry 65a of Ch. XIX) to be the case if 



W=(w 2 +ft 2 )(r 2 +s 2 ), i> = 2ran(r 2 s 2 )+2rs(w 2 ft 2 ). w = 2mn(r 2 -\-s 2 ), 



w = r 4 +6r 2 s 2 +s 4 , n = 4rs(r 2 -s 2 ). 



P. Cheluccii 26 treated Diophantus III, 7. From x+y+z = r 2 , x+y = s 2 , 

 x+z = t 2 , y+z = v 2 follow x = t 2 -r 2 +s 2 , y = r z -t 2 , z = r 2 -s 2 , 2r 2 -t 2 -s 2 = v 2 . 

 Set t = r-m, s = r-n. Then r = (y 2 +m 2 +n 2 )/(2m+2n). 



L. Euler 27 treated the problem to make x+a, x-\-b, x+c all squares. 

 Set x = z~ a, z = p/q, b a = m, c a = n. Then p 2 +mq 2 and p 2 +nq 2 are 

 to be squares.* This is impossible if m = n=f 2 or 2/ 2 , and if m = 1, n = 2. 

 Several solutions are found when m 2, n = Q. In 213-8, pp. 264-271 

 (Opera, 446-9), he made x-\-a, x-\-b squares, also a-\-x, ax. 



Euler 28 treated the problem to make xy, xz, yz all squares. Let 

 y = x p 2 , z = x q 2 , and p 2 +r 2 = q 2 , whence yz = r 2 , y-{-z = 2x p 2 q 2 . 

 Equate the last sum to t 2 , whence 2x t 2 -\-p 2 -}-q 2 . It remains to make 

 x-\-y = t 2 +q 2 and x+z = t 2 +p 2 both squares. To satisfy p 2 -\-r 2 = q 2 , take 

 p = a 2 b 2 , r 2db, q = a 2 -\-b 2 . To make ^ 2 +g 2 and 2 +p 2 squares, viz., 

 ^ 2 +a 4 +6 4 db2a 2 6 2 =D, it suffices to make 2 +a 4 +6 4 = c 2 +d 2 , 2a?b 2 = 2cd ) 

 which are satisfied if a =fh, b = gk, c =f 2 g 2 , d = h 2 k 2 , and 



(2) * 2 =(/ 4 -& 4 )(<7 4 -/i 4 ). 



By means of a table of values of m 4 n 4 for m^!5, n^Q, n<m, he found 

 the solutions 520 2 =(3 4 -2 4 )(9 4 -7 4 ) and 975 2 =(3 4 -2 4 )(ll 4 -2 4 ) of (2) 

 and hence 



x = 434657, y = 420968, 2 = 150568, 



x = 2843458, y = 2040642, 2 = 1761858. 

 J. L. Lagrange 29 treated a-\-bx = t 2 , c+dx = u 2 by eliminating x; thus 



(dt) 2 = dbu 2 +(ad-bc)d ) 



the second member being made a square in the usual way. To make 



t 2 , cx-}-dy = u 2 , 



24 Math. Quest, proposed in Ladies' Diary, 2, 1817, 19-22. 



25 New Series of Math. Repository (ed., T. Leybourn), 3, 1814, I, 1G3, Quest. 310. 



26 Institutions analyticae, Viennae, 1761, 135. 



27 Algebra, St. Petersburg, 2, 1770, 223; French transl., Lyon, 2, 1774, pp. 281-5. Opera 



omnia, (1), I, 454-6. Cf. Haentzschel 163 of Ch. XXII and paper 82 below. 

 * Euler's further discussion will be given under concordant forms, Ch. XVI. 



28 Algebra, 2, 1770, 235; 2, 1774, pp. 314-9. Opera Omnia, (1), I, 470-3. Same problem 



in papers 12, 14, 17, 18, 22, 23, 24, 30, 33, 34, 57, 74, 85, 89. See papers 40-45 of Ch. 

 XIX. 



"Addition VI, arts. 62-63, to Euler's Algebra, 2, 1774, 557-561. Euler's Opera Omnia, 

 (1), I, 595-7. Oeuvres de Lagrange, VII, 115-7. 



