CHAP. 1] POLYGONAL, PYRAMIDAL AND FIGUEATE NUMBERS. 23 



while 5, 7, 9, 13, 21, 11 terms are needed to express every number as a sum 

 of tetrahedral, octahedral, cubic, icosahedral, dodecahedral, and squares of 

 triangular, numbers. Legendre had proved that Sn + 3 is a sum of three 

 odd squares, each being 8 A + 1. Pollock gave the generalization that, 

 if F x is any figurate number of order x, SF X + 3 is a sum of 3 or 3 + 8, , 

 or 3 + 8n terms of a series whose general term is SF y + 1. 



V. Bouniakowsky 118 employed (1) and (9) of Vol. 1, Ch. X, of this 

 History, to prove that every odd pentagonal number can be expressed as a 

 sum of another pentagonal number and either a square or the double of a 

 square; every odd square not a triangular number is a sum of double a 

 triangular number and either a square or the double of a square. Similarly, 



(1, 2)a 2 = A x + (1, 2)w 2 , A x = A, + (1, 2)w 2 , 

 the factor (1, 2) denoting 1 or 2. 



F. Pollock 119 stated without proof that any integer between two con- 

 secutive triangular numbers is the sum of four triangular numbers the sum 

 of whose bases is constant. 



J. B. Sturm 120 gave the relations 



(2n + I) 2 + (4A) 2 = (4An + I) 2 , 



(2n + l) 2 (2m + I) 2 + (4 A B - 4A TO ) 2 = (4A + 4A m + I) 2 . 

 V. A. Lebesgue 121 gave two proofs of the final theorem under Wallis. 44 

 J. Liouville 122 proved readily that the only forms a A + 6 A' + cA" 

 which represent all numbers, where the A's are triangular numbers and 

 a, 6, c are positive integers, are A + A' + cA" (c = 1, 2, 4, 5) and 

 A + 2 A' + d A" (d = 2, 3, 4). That conversely each of these seven forms 

 represents all numbers is proved by use of Legendre's theorem that a 

 number = 1, 2, 3, 5, 6 (mod 8) is a SI. The case c = 1 was treated by 

 Gauss. 82 Next, 



2(2n + 1) = 4w 2 + (2t + I) 2 + (2z + I) 2 



8n + 4 = (2u + 2t + I) 2 + (2t - 2u + I) 2 + 2(2z + I) 2 , 



n = Au+t + At-u + 2 A*, 

 proves the case c = 2. Next, 



8n + 6 = (2x + I) 2 + (2y + I) 2 + 4(2z + I) 2 , n = A* + A y + 4 A,, 

 Sn + 5 = (2x + I) 2 + 4(2s + I) 2 + 16i 2 



= (2x + I) 2 + 2(2s + 1 + 2<) 2 + 2(2s + 1 - 2/) 2 , 

 n = A x + 2As+t + 2A s - t , 

 or case d = 2. Next, as shown by Gauss, 



Sn + 7 = D + D + 2D = (2x + I) 2 + 4(2z + I) 2 + 2(2y + I) 2 , 



n = A x + 2 A v + 4A Z . 

 The proofs for the remaining cases c = 5 and d = 3 are longer. 



118 M6m. Acad. Sc. St. P<tersbourg, (6), 5, 1853, 303-322. 



119 Phil. Trans. Roy. Soc. London, 144, 1854, 311. 



120 Archiv Math. Phys., 33, 1859, 92-3. 



121 Introduction a la theorie des nombres, Paris, 1862, 17-20 (26-8). 



122 Jour, de Math., (2), 7, 1862, 407; 8, 1863, 73. 



