CHAP. I] POLYGONAL, PYRAMIDAL AND FIGURATE NUMBERS. 19 



Cauchy 95 denoted the xth polygonal number of order m + 2 by 



x(x - 1) 

 x m = - -- m + x, 



2i 



and proved that if A, B, , F are integers, not divisible by the odd prime 

 p, there exist integers xi, , x n , such that 



AiT + Bx m 2 + + Ex': + F^Q (mod p), 



where n = m if m is even, and n = 2m if m is odd. The case m = 2 shows 

 that there exist integral solutions of [Lagrange, 9 etc., of Ch. VIII] 



Axl + Bxl + C = Q (modp). 



L. M. P. Coste 96 showed that the problem to make two integral functions 

 of one variable equal to polygonal numbers of a given order can be reduced 

 to the problem to make two functions equal to squares. Let 



P(Z) = (pZ 2 + qZ}/2, A = Az* + A'z + P(o), /, = Bz 2 + B'z + P(6) . 



Then to make /i and / 2 equal to numbers P(Z), take Z = az + a and 

 Z = $z + 6 in the respective cases. We obtain a quadratic equation for a 

 and one for , each linear in z. Solving for a. and /?, we require that the 

 quantities under the radical signs be squares, viz., Spfi + q 2 = D, 

 8p/2 + q 2 = D. Next, if / t and / 2 are of the form 2a 2 pz 2 + Az + A', use 

 Z = 2az + a. We can make two quadratic functions equal to P(Z) if a 

 particular solution is known. 



Several solvers 97 readily found two pentagonal numbers P x = (3z 2 x)/2 

 and P y whose sum and difference are triangular by solving 



S(P X Py) + 1 = D. 



A. M. Legendre 98 concluded from the formula 

 (6) (1 + q + q* + q + q" + O 4 = + + + '" 



that every integer N is a sum of four triangular numbers in a(2N +1) ways, 

 where a(k) denotes the sum of the divisors of k. He gave an identity which 

 shows the number of ways N is a sum of eight triangular numbers. 

 Cauchy 99 gave (6); it was attributed to Jacobi by Bouniakowsky (see 

 Vol. I, Ch. X 12 - 19 of this History). Cf. Plana. 123 



Several 100 found numbers > 1 which are simultaneously triangular, 

 pentagonal and hexagonal. Let \m(m + 1) = l(3w 2 n) 2p 2 p. 

 Then m = 2p - 1, n = (1 + fl)/6, where R 2 = 48p 2 - 24p + 1. Thus 

 1 + R = 6kp, whence p = (2 - fc)/(4 ~ 3A; 2 ). Take k = b/a. Then p is 

 integral if 4a 2 - 36 2 = 1. By the continued fraction for V3, we get 



98 Jour, de 1'Ecole Polyt., Cah. 16, Vol. 9, 1813, 116-123; Oeuvres, (2), I, 59-63. 



98 Annales de Math, (ed., Gergonne), 10, 1819-20, 101-122. 



87 The Gentleman's Math. Companion, London, 5, No. 30, 1827, 558-9. 



98 TraitS des fonctions elliptiques, 3, 1828, 133-4. 



89 Comptes Rendus Paris, 17, 1843, 572; Oeuvres, (1), VIII, 64. 



100 Ladies' Diary, 1828, 36-7, Quest. 1468. 



