CHAP, xxi] CUBE OF SUM PLUS ANY ONE A CUBE. 611 



A. B. Evans 424 found three positive numbers whose sum is unity such 

 that each plus unity is a cube. Take a : -z 3 1 as the numbers. Then 

 X 3 2a 2 i =4. Set p = s+a 3 , %r A = a\-}-al s 3 . Eliminating the a*, we get 



r~ , 12r 3 s 



if V = 



x 3 4 V 4r 2 / ' ~4r 3 -s 3 ' 



Then z=(4r 3 -s 3 )/(4r 4 +2rs 3 ). For r = 9, s = 5, the condition a*+al = : 

 is satisfied if ai = 1404/133, a 2 = 1637/133. Cf. papers 426, 428. 



D. S. Hart 425 found N (N^5) numbers such that if each be subtracted 

 from the cube of their sum s the remainder is a cube. For 3 numbers 



ijo ^^i i ^ ^ mi o i Q i t rt Q 



Xn i fy I OT O" _. O* 'YY\^ O" ^^ Q t ~~ >!" O" M_._ ^ Tl" I M OT^ -^" I , /vi _, I __, /nu -c r" _ -_. o 



j i/> ^j 1CL o *i/ //' } i/ ftjO iC /^* -^ Alcll //fr \^il {^ p ' Oo Oj 



which is satisfied if 



These give answers involving the least numbers found to date. For AT = 4, 

 m 3 +w 3 +p 3 +5 3 = 4s 3 s, which is satisfied if s = 5/9, m = 3/27, n = 5/27, 

 p = 6/27, q = 13/27; the desired numbers are the ratios of 3348, 3250, 3159, 

 1178 to 27 3 . For N = 5, take s = f, and m, - , r to be the ratios of 1, 3, 4, 

 5, 8 to 18. 



R. Davis 426 divided unity into three parts such that each increased by 

 unity is a cube. He and D. S. Hart (p. 133) treated Diophantus V, 20. 



S. Tebay, 427 to make a 3 Xi a cube where a = Saj,-, and hence na 3 a a 

 sum of n cubes, would express n or 2 as a sum of n cubes, the roots of 

 three of which are ms, m t, s+tm. Let Hn+m z be the negative of 

 the sum of the remaining cubes. Then a~ 2 = H-\-3st(2m s t). Equate 

 the last product to 9r 2 s 2 i 2 , thus determining t. Then 



Hence s and t are found rationally in terms of r, m, H. He 428 expressed 2 

 as a sum of three rational cubes. But 3s 3 s = 2 if s = l. Hence, as by 

 Hart, 425 we have three numbers whose sum is unity and such that unity 

 exceeds each by a cube. He tested eleven sums of three cubes by the 

 method of Hart, 425 but found no answer in quite so small numbers as Hart's, 

 his smallest answer being 13/49, 17/64, 351/(49-64), with the sum 9/14. 



A. Holm 429 treated Diophantus V, 19 by starting with Diophantus' 

 formula 



To find positive solutions of 3 Set? = D , take a i = 5/6, a 2 = f x, o 3 = | + x. 

 Then 



97 1 / 3 V 36r+7 



424 Math. Quest. Educ. Times, 25, 1876, 31. Cf. Strong 61 and Lenhart 62 of Ch. XXV. 

 426 Ibid., 26, 1876, 66-8. 



426 Math. Visitor, 1, 1880, 107. 



427 Math. Quest. Educ. Times, 38, 1883, 81-2. 



428 Ibid., 101-3. 



429 Math. Quest, Educ. Times, (2), 9, 1906, 98. 



