G10 HISTORY OF THE THEORY OF NUMBERS. [CHAP, xxi 



By trial, G = 37 2 if a = 5, b = S, c = 9, u = l. Setting b = 3q- a, we see that 

 G becomes a quadratic in a, and G= (m 3qa) 2 determines a rationally. 



M. Noble 420 gave a note on the history of Diophantus V, 19, citing 

 papers reported on above. He noted that one solution leads to an infinitude. 

 For, if 3-Sa- = a 2 , then 3-2(a i +g i x)* = (a+fx') z , provided 



We may take 3A+2a/=0, z=(-/ 2 -3)/C. He also gave the following 

 solution. Let x, y, z be the desired numbers and s their sum. Then 

 x = s 3 a 3 , y = s 3 6 3 , z = s 3 c 3 . Thus 



(1) s = 3s 3 -a 3 -6 3 -c 3 . 



Take s = u 2 v, a = (p-\-q)v, b = (u z q)v, c=(u i p)v. Then (1) requires that 



Set E= (u? -\-pmfn) 2 . We get p and hence 



Equating F to the square of u 3 k-\-qr/e, we get q. Then evidently 



v = - -, x- [u*- 

 eku A -\-rq 



Wm. Lenhart 421 found n numbers Xi such that if each be added to the 

 cube of their sum s the sum shall be a cube a*. Thus s+ns 3 = Saf. Take 

 s = l/r. Then r 2 +n = Z(ra;) 3 . But m another paper, Lenhart 62 of Ch. 

 XXV, he showed how to express a number (here r--\-ri) as a sum of cubes. 

 Again, to find n numbers Xi such that if each be subtracted from the cube 

 of their sum s the remainder shall be a cube &\, we have ns 3 s Sj8< . Take 

 s = r/t. Then r(nr 2 < 2 ) = 2(<|8 t -) 3 . If r = t, the problem is to find n cubes, 

 each <1, whose sum is n 1. It was discussed in the paper cited. Here 

 let t>r, t being prime to r. The following tentative process was used. 

 From nr 2 subtract in turn the terms of a decreasing series of squares prime 

 to r and beginning with the first square <nr 2 and ending with the square 

 just >r 2 ; Multiply each remainder by r and seek (as in the paper cited) a 

 separation of the product into cubes (/3;) 3 . For n = 4, take r = 12, =19; 

 2580 = a+6, a = 1241 = 9 3 +8 3 , 6 = 1339 = 2 3 +11 3 . Using his table of sums 

 of two cubes, he found various answers for n = 3 and one for n = 5. 



S. Bills, 422 to find Xi=(l-a^ t s = 2x f , would solve n-a\ ----- al = W 

 by taking arbitrary values for k, a 4 , , a n and using the theorem that any 

 number is a sum of three rational squares. Similarly, 423 to find Xi = (a< l)s 3 , 

 we have Sa? n = l/s 2 ; set K = l/s and assign arbitrary values to o 3 , , a n 

 and solve o 3 + a 3 . + d = K 2 , where d = al H ----- f- a* n. Put 



then K 2 -9v* = 27l+d=Jg. Taking K+3y=/, K-3v = g, we get # and 



420 Leybourn's Math. Quest, proposed in Ladies' Diary, 1, 1817, 52-62. 



421 Math. Miscellany, New York, 1, 1836, 263-7. 



422 Math. Quest. Educ. Times, 22, 1875, 71. 



423 Ibid., 24, 1876, 52-3. 



