68 HlSTOKY OF THE THEOEY OF NUMBERS. [CHAP. II 



proved in two ways, one by use of a geometric process communicated to 

 him by Lucas: Given one point on the line ax + by = n with integral 

 coordinates == 0, it is easy to find all such points. If M is the point with the 

 maximum abscissa, we get a second point M' by subtracting b from the 

 abscissa of M and adding a to the ordinate of M. From M' we obtain 

 similarly a new point, etc. 



Cesaro stated and N. Goffart 129 proved that the total number of integral 

 solutions ^ of 



x + 4y = 3(n - 1), 4x + 9y = 5(n - 2), 9x + Wy = 7(n - 3), 



is n. 



J. Gillet 130 stated that the sum of the numbers of solutions of 



p m x + (p + l} m y = {(p + T) m p m }n p m (p = 1, , n) 



is n, a generalization of the theorems by Cesaro 126 and Catalan. 127 



E. Lucas 131 proved Catalan's 123 result and added the remark that there 

 are %(a 1)(6 1) values of his n' for which ax + by = n' has no solutions 

 ^ 0. In the continued fraction for a/b, let a/ft be the convergent of rank 

 n 1 immediately preceding a/b. Then, writing r for n', we have the 

 solution x = ( I) n rf3, y = ( l} n ra of ax + by = r. The sum of 

 the squares of the values x = X Q + bt, y y at, giving the general 

 solution, is a minimum for t = s/(a 2 + 6 2 ), where s = ( \} n ~ l (aa + b/3)r. 

 Let pi be the least positive remainder and p 2 the greatest negative 

 remainder when s is divided by k = a 2 + 6 2 . Then the sets of minimum 

 solutions are given by 



kxi = ar bpi, kyi = br + api, kx 2 = ar + bp 2 , ky z = br ap z . 



In only one of the sets are the unknowns ^ 0. Hence ax + by = r is 

 solvable in integers ^ if and only if one of ar bpi and br ap% is not 

 negative. 



E. Catalan 132 showed by an example that Lucas' last method requires 

 long computations. He noted (ibid., 241-3) that, if w(n) denotes the 

 number of integral solutions ^ of px + qy = n, 



2 1 

 2 2 co(2) + - - + 2*-*-o>(p ff - p - <?) = 



A. S. Werebrusow 133 noted that co = (n b/3 aa)/ab, if /3 is the least 

 positive y, and a the greatest negative x. 



L. Salkin 134 employed the argument of Catalan 123 to show that w = q or 

 q + 1, according as d ^= d' or d > d", where, if I, m is one set of solutions, 

 d = l/b-- p/6], d' = m/a-- 



129 Nouv. Ann. Math., (3), 3, 1884, 399, 539-40. 



130 Mathesis, 6, 1886, 32. 



131 Mathesis, 10, 1890, 129-132; Thdorie des nombres, 1891, 479-484; Jour, de math. sp6c., 



1886, 20-22. 

 * Mathesis, 10, 1890, 197-9. 



133 Spaczinski's Bote Math., Odessa, 1901, Nos. 298, 299. 



134 Mathesis, (3), 2, 1902,|107-9. 



