8 HISTORY OF THE THEORY OF NUMBERS. [CHAP. I 



the figurate numbers /, fl, f\, f r 5 , so that his work amounts to a verification 

 of cases of 



Frans van Schooten 45 quoted three of Bachet's rules, proving one. 



On certain hexagons whose sum is a cube, see Frenicle 6 of Ch. XXI. 



Fermat 46 proposed that Broun cker and Wallis find a proof of the pro- 

 position (which he himself could prove): There is no triangular number, 

 other than unity, which is a biquadrate. 



Diophantus, IV, 44, desired three numbers which if multiplied in turn 

 by their sum give a triangular number, a square, and a cube. Let the sum 

 be x 2 . Then the numbers are a(a + l)/(2z 2 ), /3 2 /a; 2 , 7 3 /z 2 . Thus 



A a + /3 2 + 7 3 = * 4 . 

 Take = z 2 - 1. Then A a = 2z 2 - - T 3 - 1. But 



8A a + 1 = (2a + I) 2 = 16z 2 - 87* - 7 = (4x - 5) 2 , 



if x = (87 3 + 5 2 + 7)/(85). Take 7 = 2, 5 = 1; then x = 9 and the 

 desired numbers are 153/81, 6400/81, 8/81. 



Bachet convinced himself by trial that 6 must be unity in order that 

 a = (8 7 3 + 7 - 5 2 - 25)/(4S) be integral. 



Fermat remarked that "Bachet's conclusion is not rigorous. Indeed, 

 let 7 be any number of the form 3n + 1, say 7 = 7. To make 



2z 2 - 7 3 - 1 = A 



and hence 16x 2 8-7 3 7 = D, we may take the latter to be the square 

 of 4x 3 [whence x = 115, 5 = 3} Nothing prevents us from gen- 

 eralizing the method, taking instead of 3 any odd number and making a 

 suitable choice of 7." 



G. Loria 47 remarked that the solution becomes evident if we replace 

 x* by x] the problem did not require that the sum of the numbers be a 

 square. 



Bachet 32 (p. 274) proposed the problem to find five numbers which if 

 multiplied in turn by their sum give a triangular number A, a square, a 

 cube, a pentagonal number, and a biquadrate. The sum of the latter 

 shall be x 4 . Let the square be (x 2 I) 2 , the cube 8, the pentagonal number 

 5, and the biquadrate 1. Then A = 2z 2 15. Thus 



8A + 1 = 16z 2 - 119 = D, 

 say (4x I) 2 . Hence x = 15. 



Rene" F. de Sluse 48 (1622-1685) employed the triangular number q, the 

 square 6 2 and cube z 3 . Then q + & 2 + z 3 = D = (6 + n) 2 , whence 



6 = (q + z 3 - n 2 )/(2n). 



Exercitationum Math., 1657, Lib. V, 442-5. 



46 Oeuvres, III, 317, letter to Digby, June, 1658. 



47 Le scienze esatte nelT antica Grecia, Libro V, 138. 



"Renati Francisci Slusii, Mesolabum, , accessit pars altera de analyai et miscellanea, 

 Leodii Eburonum, 1668, 175. 



