38 HlSTOKY OF THE THEORY OF NUMBERS. [CHAP. I 



G. Picou, 9, 1902, 115, noted that a 2 - 6(6 + l)/2 = 2 2n for 



a = 2 n+1 + 2 B 1, 6 = 2"+ 2 +1 or 2"+ 2 - 2. 

 H. Brocard, 10, 1903, 24-6, noted the solution 



a = (9- 2" - 2)/7, 6 = 8(2- - l)/7. 



P. F. Teilhet, 10, 1903, 240-1, gave a somewhat general discussion. 



That 1 + 6 A + cube 4= 0, see 10, 1903, 97, 197. 



P. Jolivald, 12, 1905, 16, 152, gave an erroneous proof that unity is the 

 only number simultaneously a A, square and hexagon. As noted by M. 

 Rignaux, 24, 1917, 80-1, a hexagonal number r(2r 1) is triangular, so 

 that we have only to solve r(2r 1) = ?/ 2 , whence Sy 2 + 1 = D, whose 

 solution is known. 



A product of 3 consecutive A's may be a square, 11, 1904, 158. 



H. B. Mathieu, 16, 1909, 34, gave identities showing that the square of 

 any number =f= 1 [4, 16], and not a multiple of 3, is a sum of a A and a 

 square, each not zero [three A's]. 



A. Arnaudeau, 18, 1911, 132, deposited with the library of the Institute 

 of France the manuscript of his unpublished table of triangular numbers. 



A. G<rardin, 1911, 205-7, gave solutions of x* + tf + z 4 = 2T*/P, where 

 T and t are triangular numbers A. He cited, 273, Fuss' 67 note giving 9z+5, 

 9x + 8 as linear forms of numbers not a sum of two A's. 



To decompose (n + I) 5 - n 5 into three A's see 19, 1912, 37, 104-5. 

 For 



(x + I) 3 + (x + 2) 3 + - - + (x + m) 3 = Al +m - A, 



see 19, 1912, 114. L. Aubry, 19, 1912, 231; 20, 1913, 108, noted that 

 A, - Aj = z 3 for x, y = (8m 4 =b 12m 3 - 4m 2 - l)/3. A. S. Monteiro, 

 20, 1913, 18-20, obtained solutions from the fact that the sum of the cubes 

 of any number of consecutive integers equals the difference of the squares 

 of two A's. 



R. Niewiadomski, 238 20, 1913, 5-6, gave many algebraic identities be- 

 tween polygonal numbers, also expressions for n k , n 3 + 1, n 3 + (n + I) 3 , 

 etc., as polygonal numbers. 



U. Alemtejano (a pseudonym), 21, 1914, 169, stated that if 4m + 1 is 

 a sum of two squares, m is a sum of two A's, and conversely, since 



4( A n + A) + 1 = (n + a + I) 2 + (n - a) 2 . 



Also, 9 is the only number 4 A n + 5 which is a square of a prime and not a 

 sum of two squares. Again, Az n + a + A a -i n = n 2 + (n + a) 2 . Proofs 

 by L. Aubry, 22, 1915, 69. Alemtejano, 22, 1915, 8, gave 



{4(An + A ) + I} 2 = (2a + l) 2 (2n + I) 2 + {4(A n - A a )} 2 . 



Several, 22, 1915, 167-8, proved that every square is expressible in the form 

 AM 2 A v in an infinitude of ways. On the last digits of A, see 22, 1915, 

 235-6. 



238 Also in Wiadomsci Mat., Warsaw, 17, 1913, 91-98. 



