CHAP. XXI] SUM OR DIFFERENCE OF TWO CUBES A SQUARE. 579 



To obtain relatively prime integers x, y, when p, q are integers, we must 

 employ fractional values for n. To obviate this, Euler gave a second 

 method. The factors x+y, x 2 xy+y 2 have the g.c.d. 1 or 3. In the first 

 case, he put the second factor equal to the square of p 2 pq-\-q 2 and stated 

 that x = p 2 2pq, y = p 2 q 2 . The upper sign is excluded since 



For the lower sign, x+y = (p+q)~ 3p 2 = D if 



p = 2mn, q = 3m 2 2mn-\-n 2 , 

 z = 4mn(3ra 2 3mn+n 2 ), y = (m n)(3ra 



In the second case, x 2 xy+y 2 = 3(p 2 pq+q 2 ) 2 , (x+?/)/3=D. As the 

 three subcases lead to equivalent results, consider the case 



x = 2p 2 -2pq-q 2 , y = p 2 -4pq+q 2 , (x+y)/3 = p 2 -2pq=D. 

 The last condition is satisfied if p = 2rn?, q = m?n 2 , whence 

 z = 3ra 4 +6raV-?i 4 , y= -3m 4 +6m 2 n 2 +n*. 



Euler 226 noted the examples l+2 3 = 3 2 , 8 3 -7 3 = 13 2 , 37 3 +H 3 = 228 2 , 

 65 3 +56 3 = 671 2 , 71 3 -23 3 = 58S 2 , 74 3 -47 3 = 549 2 . 



Several 227 found that the difference of 7 3 and 8 3 is a square by considering 

 z 3 , (z+1) 3 , and, by use of tables of cubes, found that this pair and 7 3 , 14 3 

 give the least solutions. 



C. H. Fuchs 228 discussed x*+y* = az 2 . Let x, y, z have no common 

 factor, a no square factor. If x or y is even, set x-}-y = p, x y = q. Then 

 p(p 2 +3q 2 } =4a2 2 . If p is not a multiple of 3, 



p = at 2 , p 2 + 3q 2 = 4/fti 2 , a/3 = a. 

 Since ft is a divisor of p 2 -\-3q 2 , it is of that form. Thus 4/3 = /r+3*' 2 . Also 



By use of V^, he got 

 (1) p = M (s t2 -37 ? 2 )-6^r 7 , 3 = Kr-3r 



The case p = 3P is similar. For xy odd, set 2p=x-\-y, 2q=xy. One 

 of the three cases has p = 2p', a odd. Then p = 2od 2 , p 2 +3q 2 = (3u 2 . He 

 again got (1). 



R. Hoppe 229 obtained the general solution of z 3 -f-?/ 3 = 2 2 in relatively 

 prime integers by setting pq = z 2 , p = x+y, q=(x+y}(x 2y)+3y 2 , where 

 p and q have the greatest common factor 1 or 3. In the first case all solu- 

 tions are given by 



2 z = a(a 3 -86 3 ), 2 ?/ = 4&(a 3 +6 3 ), 3 z = a 6 +20a 3 & 3 -86 6 , 



where a is odd, and = 3 or 1 according as 3 is or is not a divisor of a +6. 

 Second, if p, q have the factor 3, the solutions are [Euler 225 ] 



226 Opera postuma, 1, 1862, 241. 



227 Ladies' Diary, 1812, 35, Quest. 1227; Leybourn's M. Quest. L. D., 4, 1817, 149. 



228 De Formula z 3 +z/ 3 = az 2 , Diss. Vratislaviae, 1847, 33 pp. 



229 Zeitschrift Math. Phys., 4, 1859, 304-5. 



