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



(2o, 6) = (2, 1), (26, 15), (362, 209), (5042, 2911), , whence p = 1, 143, 

 27693, -, so that answers are 1, 40755, 1533776801, 



J. Whitley 101 found pairs of pentagonal numbers p, q whose sum and 

 difference are pentagonal. The conditions are 



24(? + 1 = y 2 , 24(p + q) + 1 = z 2 , 



Hence z 2 = x 2 + y 2 - 1, y 2 = x 2 - y 2 + 1. Letz 2 = n 2 + m 2 , y 2 = 2nm + 1. 

 Then z = n + m, v = n m. Take n = r 2 s 2 , m = 2rs, whence 

 x = r 2 + s 2 . There remains the condition 



4rs(r 2 - - s 2 ) + 1 = D = y 2 . 



This is said to hold if r = K0 5 - <), s = i(4> 5 30), which lead to larger 

 numbers than those found by trial, using (r, s) = (3, 2), (6, 1), (8, 5), 

 (13, 2), (13, 8), (19, 14). [But the resulting numbers p = 7, 37, 330, 

 are not pentagonal.] See Gill. 108 



C. G. J. Jacobi 22 of Ch. VII gave in 1829 the result 



+ 00 ] 3 oo 



= Z(-l) n (2n+l)z< 



m= oo 



the exponents on the left being pentagonal numbers for m negative, and 

 those on the right triangular. Polygonal numbers appear incidentally in 

 Jacobi's paper of 1848 [see Ch. III]. 



J. Huntington, 102 given a pentagonal number P = r(3r l)/2 of n 

 digits, found another number p also of n digits such that if p is prefixed to 

 P there results a pentagonal number. Let x be the root of the latter. 

 Then shall 10 n p + P = x(3x - l)/2. Taking p = x - r, we get 



x = |(2 -10 B + 1) ~ r. 



For example, let r = 1; then n = 1, x = Q, p = 5 and 51 is pentagonal. 



A. Cauchy 103 defined triangular and pyramidal numbers as binomial 

 coefficients. 



J. Baines 104 found two squares x 2 , y 2 whose sum and difference are 

 hexagonal. Take 8(z 2 - - y 2 ) + 1 = (2(x + y) 1 } 2 , whence x = 3y db 1. 

 Then 8(z 2 + y 2 ) + 1 = 80?/ 2 48y + 9 = (ny 3) 2 determines y. 



A. Bernerie 105 gave a table of triangular numbers. 



A. Casinelli 106 noted that every triangular number is of one of the three 

 forms 



(9m 2 + 3m) /2 = A m -i + 2 A 2 m, (9m 2 + 9m + 2)/2 = Am+ A 2 m+ A 2 m+i, 



(9m 2 + 15m + 6)/2 = A OT+ i + 2 A 2m+ i, 



also a sum of four A's, and hence a sum of any number of A's. By adding 



101 Ladies' Diary, 1829, 39-40, Quest. 1489. 



102 Ladies' Diary, 1832, 36-7, Quest. 1530. 



103 Re'sumc's Analyt., Turin, 1, 1833, 5. 



104 The Gentleman's Diary, or Math. Repository, London, 1835, 33, Quest. 1320. 



105 Nouv. table des triangulaires, Bordeaux, 1835. 



loe N ov i Comm. Acad. Sc. Inst. Bononiensis, 2, 1836, 415-34. 



