CHAP. I] POLYGONAL, PYRAMIDAL AND FIGURATE NUMBERS. 21 



the first two or the second and third equations, we get 



(3m + I) 2 = m 2 + (2m + I) 2 + 2A 2 , 

 (3m + 2) 2 = (m + I) 2 + 2A 2m+ i + (2m + I) 2 . 



Also, (3m + 3) 2 = (m + I) 2 + 2 (2m + 2) 2 . Hence every square is a sum 

 of three squares or a sum of two squares and two A's. Further, every A 

 is a sum of a square and two equal A's. Next, 



A m + An + mn = A TO+n , A m + A n + (m + l)(w + 1) = A m +n+i, 



and similarly for three or more A's. Also, A m H- A n m(n + 1) = A n -m. 

 C. Gill 107 found numbers both m-gonal and n-gonal, and the generaliza- 

 tion 



T = ax z - a'x = by 2 - b'y, 



where a, a', b, b' are given integers with no common factor. Take 



ax - a' = yp/q, x = (by - b')qfp, 

 so that 



x = q(b'p + a'bq)/N, y = q(a'p + b'aq}/N, - N = p 2 - abq 2 . 

 Let p', q' give a particular solution of the last equation such that 



A = (a'p r + ab'q')/N, B = (Vp f + ba'q')/N 

 are integers. Take p = p't + abq'u, q = q't + p'u. Then 



where F = P - abu 2 , and a; = g(5^ + Abu)/F, y = g(A^ + Bau^/F. From 

 the initial solution to = 1, MO = 0, of F = 1, we get as usual the solution 



It remains only to find a solution p', q' of p 2 abq 2 = N. While one 

 may employ the continued fraction for Va6, it suffices for our initial problem 

 to note the solution p' = a a', q' = 1, for the case a a' = b b'; 

 then N = ab' + ba' a'b', A = B = 1. First, if m and n are both odd, 

 we may take 



a = m 2, a' = m 4, b = n 2, b' = n 4, 

 which have no common factor. Then a a' = b b' = 2. For P t = IT 7 ,-, 

 Po = 1, Pi = 2UVi - P t _ 2 + (2d - l)(fc - 1), 



(m + n 4}(mn 2m 2n + 8) 

 16(m -~2)(w -^T 



But if m and n are both even, take a = \m 1, a' = fm 2, 6 = |n 1, 

 6' = n 2, whence o a' = 6 6' = 1, and P, = I 7 ,- satisfies the same 

 recursion formula. Also, 



PX = \(U + 1) + d(U - 1) + -mnui, 



e 



107 Math. Miscellany, Flushing, N. Y., 1, 1836, 220-5. 



