CHAP, xxi] Two EQUAL SUMS OF Two CUBES. 551 



Using the same sides for (2); sides A-D, D~A/B 2 -B for (3), he got 



B(B 3 +2D 3 ) 

 (2) x*-y*=** 



3 - 3 



B 3 ) 



C. G. Bachet, 39 in his commentary on Diophantus IV, 2 (to solve 

 x y = g, x 3 y 3 = 1i), gave Vieta's results (l)-(3). He was able to express 

 the difference of two given cubes as a sum of two positive cubes only when 

 the greater of the given cubes exceeds the double of the smaller. 



A. Girard 39 " noted that, if D 3 >B 3 in (1), we first apply (3) repeatedly 

 until we obtain two cubes the smaller of which is less than one-half the 

 larger, and then use (1). 



Fermat 40 noted that in the case B 3 < 2D 3 , expressly excluded by Bachet, 

 we can make B 3 D 3 a sum of two positive cubes. Let, for example, 

 B = 5, D = 4. By Vieta's formula (3), we get 



P;3 _ 4.3 _ ( 2jL8\3 _ f_5\3 

 - I 63 j (63) - 



Of the new cubes, the first exceeds the double of the second. Hence their 

 difference is a sum of two cubes by (1). Thus 5 3 -4 3 is the sum of two posi- 

 tive cubes, " which would doubtless astonish Bachet." Further, if we 

 employ the three formulas in succession, and repeat the operations in- 

 definitely, we obtain an infinitude of pairs of cubes satisfying the same 

 conditions ; for, from the two cubes whose sum equals the difference of the 

 given cubes, we can find by (2) two new cubes whose difference equals 

 the sum of our two cubes and hence equals the difference of the two original 

 cubes; from this new difference of two cubes we pass to a sum of two cubes, 

 and so on indefinitely. The condition B 3 <2D 3 imposed by Bachet on (3) 

 is not necessary; being given the cubes 8 and 1, we can find two new cubes 

 with the same difference. Bachet would doubtless say that this is impos- 

 sible. Nevertheless I have found that 41 



/J_2.6_5\3 fl256N3_C 1 

 ( -T83 ) -\TTWS~) ' - 1 - 



Further, after what precedes, I solve happily the problem (not known by 

 Bachet) : To separate the sum of two cubes into two new cubes, and indeed 

 in an infinitude of ways. Thus to find two cubes whose sum is 8 + 1, 

 I first seek by (2) two cubes 8000/343 and 4913/343 whose difference is 8+1. 

 As the double of the smaller exceeds the larger, we apply (3) and afterwards 

 (1) and obtain the solution. If we wish a second solution, we apply (2), 

 etc." 



Fermat 42 proposed as a new problem to Brouncker, Wallis and Frenicle : 

 Given a number composed of two cubes, to divide it into two other cubes. 



39 Diophanti Alex. Arith., 1621, 179-182, 324. 



390 L'arith. de Simon Stevin . . . annotations par A. Girard, Leide, 1625, 635; les Oeuvres 

 Math, de Simon Stevin de Bruges par A. Girard, 1634, 159. 



40 Oeuvres, I, 297-9; French transl., Ill, 246-8. 



" By (1), 8-1 = (4/3) 3 +(5/3) 3 . Then apply (2) for B = 5/3, Z) = 4/3. 



42 Oeuvres, II, 344, 376; letters from Fermat to Digby, Aug. 15, 1657; Apr. 7, 1658. 



