CHAP. XXI] EQUATIONS OF DEGREE THREE IN THREE UNKNOWNS. 605 



and took x = 2mn, y = m z n 2 . Then shall ?n- + kmn + n~ = D , say (m + pri) 2 , 

 which gives m/n. Take p = 3/2. Then xy is a cube if n 300. 



J. Matteson 389 solved the last problem and that with the final condition 

 replaced by the following : The sum of the roots of the squares is an eighth 

 power, the squares being a seventh, a fifth and a fourth power, and the 

 arithmetical mean of the required numbers a square. 



To find three positive integers whose sum, sum of squares, and sum of 

 cubes, are squares, A. B. Evans 390 took ax, ay, az, where a = x+y-\-z, and 

 set a(z 3 +?/ 3 +2 3 )=a 2 ( y+z) 2 , whence y 2 +y(x-\-z) 3xz = Q. The radical 

 in y is rational if x 2 -}-14xz-\-z 2 = D = (zm/nx) 2 , which holds if x = m 2 n 2 , 

 z = 2mn-\-14:n 2 . In the resulting expression for 2z 2 , set m = p 8n and 

 equate to the square of p 2 -lGpn-83n 2 . Thus p/n = 1332/83. Then 

 az = 412095790665, etc. Several solvers used I5mx, 15my, Smfa+y}, whose 

 sum of cubes is divisible by their sum. Thus a linear and two quadratic 

 functions are to be squares, which is true if m = d 2 /{23(x-\-y} }. 



D. S. Hart 391 divided unity into three positive parts whose sum of squares 

 and sum of cubes are squares by taking x/s, y/s, z/s as the parts, where 

 s = x+y+z. The conditions are satisfied if s= D, Z 2 = D, 2x 3 = D, which 

 is the preceding problem. He 392 found three numbers whose sum and sum 

 of squares are cubes, and sum of cubes a square. Let ax 3 , bx 3 , ex 3 be the 

 numbers. Their sum of cubes will equal (a; 5 ) 2 if z = 2a 3 . To make 2a 

 and 2a 2 cubes, equate their product to (a +6 c) 3 ; the roots of the resulting 

 quadratic for a are rational if + 2b 3 c - 96V + 66c 3 - 7c 4 = D . Set 6 = 2c + d. 

 The new quartic is a square if d = 35c/9 or 116c/315. 



To find three integers whose sum, product and sum of squares are all 

 squares, S. Tebay 393 used the numbers xy, x(x+y), y(x+y), while A. B. 

 Evans used xa 2 , ya~, xya 2 , with x = y+l. 



D. S. Hart 394 found three numbers, say ax, bx, ex, such that if the sum of 

 their cubes be added to or subtracted from the square of each, the sums 

 and remainders are squares. Set d = a 3 +6 3 +c 3 . Then a 2 x 2 +dx 3 = D = e 2 a: 2 , 

 a 2 x*-dx 3 = D =f 2 x 2 givex=(e--a 2 )/d = (a z -f)/d. Similarly, 6 2 z 2 cfo 3 = 0V, 

 h 2 x 2 give X = (g 2 b 2 )/d=(b 2 h 2 )/d, while C 2 x 2 dx 3 = k 2 x 2 , I 2 x 2 give 



X = (k 2 -c 2 )/d=(c 2 -l 2 )/d. 



By the numerators of x, e 2 = 2a 2 -/ 2 , g 2 = 2b 2 -h 2 , k 2 = 2c 2 -l 2 . The first is 

 satisfied if a = P 2 +Q 2 ,/=2PQ-P 2 +Q 2 . As in Diophantus V, 8, take three 

 right triangles of equal area, with the hypotenuses 49+9, 49+25, 49+64. 

 For P = 7, Q = 3, we get a = 58, f=2. Similarly, P = 7, Q = 5 give 6 = 74, 

 /i = 46; P = 8, Q = 7 give c = 113, Z = 97. Hence we get ax, etc. 



Problems solved in the American Mathematical Monthly: Three num- 

 bers the sum of whose cubes is a square and sum of squares a cube (1, 1894, 

 363). Three integers the sum of any two of which is a cube (p. 208, p. 279). 



389 Collection of Diophantine Problems, Washington (ed., Martin), 1888, pp. 5-7. 



390 Math Quest. Educ. Times, 17, 1872, 30-1. 



391 lUd., 21, 1874, 100-1. 



392 Ibid., 26, 1876, 102. 



393 Ibid., 23, 1875, 31. 



394 Math. Visitor 2, 1882, 17-18. 



