CHAP. I] POLYGONAL, PYRAMIDAL AND FIGURATE NUMBERS. 27 



E. Lucas 150 listed values of A for which xy(x + y) = Az 3 has no distinct 

 rational solutions + 0. Taking y = 1 and y = x + 1, we obtain theorems 

 on triangular numbers and numbers x(x + l)(2z + 1). 



Lucas stated and Moret-Blanc 151 proved that I 2 + + x 2 = ky 2 and 

 A* + + A* = ky 2 are impossible if k = 2, 3, 6. 



S. Roberts 152 proved by use of y 2 - 2z 2 = 1 [Euler 70 ] that the A's 

 which are squares are 



4A/2 



J. Neuberg stated and E. Cesaro 153 proved that the sum of the squares 

 of n + 1 consecutive integers, beginning with the 2nth triangular number, 

 equals the sum of the squares of the n succeeding integers, each being 

 divisible by I 2 + + n\ Cf . Dostor 75 of Ch. IX. 



E. Lionnet 154 stated that unity is the only triangular number A which 

 equals the sum of the squares of two consecutive integers; 10 is the only A 

 equal to the sum of the squares of two consecutive odd integers; when A 

 is a product of two consecutive integers of which the least is double a 

 triangular number, then 4 A + 1 (and its square root) is a sum of squares 

 of two consecutive integers. 



Moret-Blanc 155 proved the preceding theorems stated by Lionnet. 



E, Cesaro 156 noted that no triangular number ends with 2, 4, 7, 9. 



S. Re*alis 157 noted that 



+ A(3p), A(5p + 3) = A(4p+2) + A(3p+2), 

 = A (A;) + A(2ap + a), k = 2ap 2 + (2a + l)p. 



E. Lionnet 158 noted that 0, 1, 6 are the only A's whose squares are A's. 

 He stated and E. Cesaro 159 proved that there is at least one and at most 

 two A's between any two consecutive squares =1= 0; at most one square be- 

 tween two consecutive A's; if there are exactly two A's between (a + I) 2 

 and (a + 2) 2 , where a > 0, there is just one A between a 2 and (a + I) 2 , and 

 just one A between (a + 2) 2 and (a + 3) 2 . 



E. Cesaro 160 denoted by V(n) the number of the first 2n triangular 

 numbers which are relatively prime to n. Let V(ri) be the number of 

 products 1-2, 2-3, 3-4, , n(n + 1) which are prime to n. Then if v 

 is the largest odd divisor of n, 



V (n) = V(n) *Q) *(ri) = V (n) 1 V 



n n v n n 



1B Nouv. Ann. Math., (2), 17, 1878, 513. 



m/Wrf., 527; (2), 18, 1879, 470-4. 



152 Math. Quest. Educ. Times, 30, 1879, 37. 



155 Nouv. Corresp. Math., 6, 1880, 232. 



154 Nouv. Ann. Math., (2), 20, 1881, 514. 



165 Ibid., (3), 1, 1882, 357. 



188 Mathesis, 4, 1884, 70. 



167 Jour, de math. sp<c., 1884, 6. 



Nouv. Ann. Math., (3), 1, 1882, 336. Proof by H. Brocard, (3), 15, 1896, 93-6. 



Ibid., (3), 2, 1883, 432 (misprints); 5, 1886, 209-213. 



160 Annali di Mat., (2), 14, 1886-7, 150-3. 



