CHAP. I] POLYGONAL, PYRAMIDAL AND FIGURATE NUMBERS. 13 



To express every number as a sum of figurate numbers 



n + a (n + l)(n -f- 2a) (n + l)(n -f- 2)(n + 3a) 

 1> 



1-2-3-4 



at least a + 2n 2 terms are necessary. Cf. Beguelin, 72 Pollock, 117 

 Maillet. 181 - 2 



L. Euler 69 remarked that Fermat's 36 theorem that every integer is a 

 sum of m w-gonal numbers would follow if we could prove that every 

 integer occurs among the exponents in the expansion of the wth power of 

 1 -\- x + x m + x 3m ~ 3 + , whose exponents are the w-gonal numbers. 

 Fermat's theorem that every integer is a sum of three triangular numbers 

 would follow if it were shown that in 



!/{(! - )(! - xz)(l - x*z)(l - xty - - } = 1 + Pz + Qz 2 + Rz* + -, 



all integers occur as exponents of x in the series for R. 



Euler 70 found squares which are triangular or pentagonal. If A z = x 2 , 

 then y 2 = Sx 2 + 1 for y = 2z + 1. If (3z 2 - z)/2 = x 2 , y 2 = 24z 2 + 1 for 

 y = Qz - 1. If (3q 2 - q)/2 = A P , (Qq - I) 2 = 3z 2 - 2 for x = 2p + 1. 

 Special solutions of the three equations y 2 = ax 2 + 6 are found by his 

 general method of treating the latter (Ch. XII, below). 



Euler 71 admitted that he had no proof of Fermat's assertion that every 

 number is a sum of three or fewer triangular numbers and noted that 

 this is true only of whole numbers, since no one of |, f, |, |, etc., can be 

 resolved into three triangular numbers. There are no rational solutions 

 x, y, z of 



1 x 2 + x y 2 + y . z 2 + z 

 2~ 2 2 



Nicolas Beguelin 72 attempted to prove Fermat's theorem that every 

 integer is a sum of s polygonal numbers of s sides. For s = d -f- 2, the 

 latter are and 1, A = d + 2, B = 3d + 3, C = Qd + 4, D = Wd + 5, 

 , a series whose second order of differences are d. Let t be the number 

 of terms > needed to produce a given sum e. For 1 < e < A, evidently 

 t ^ A - 1. Fore = A + , where 1 ^ e ^ A - 1, t ^ A; for e = 2A + e, 

 ^ ^ d - 2, t ^ d. Next, let B < e < C. For e = B + e, 1 ^ 

 ^ A - 1, t ^ A; for e = 5 + A + e, ^ ^ A - 2, < ^ A; for the 

 "doubtful case" e = B-\-A+A 1, we replace 5 by its equal 2A + d 1 

 and have e = 4A -}- d 2, = d + 2; finally, for e = 5 + 2A + e, 



69 Novi Comm. Acad. Petrop., 14, I, 1769, 168; Comm. Arith. Coll., I, 399^00. 



70 Algebra, 2, 1770, 88-91; French transl., 2, 1774, pp. 105-9 (Vol. I, Ch. V, pp. 341-354, 



for definitions of polygonal numbers). Opera omnia, (1), I, 373-5, 159-64. 



71 Acta Eruditorum, Lips., 1773, 193; Acta Acad. Petrop., I, 2, 1775 (1772), 48; Comm. 



Arith. Coll., I, 548. 



72 Nouv. Mem. Acad. Sc. Berlin, annee 1772, 1774, 387-413. 



