CHAP. XV] LlNEAE FUNCTIONS MADE SQUARES. 453 



C. F. Kausler 47 treated the problem to divide a given number a into n 

 parts such that the sum of any n l parts shall be a square. [The treat- 

 ment by Diophantus, V, 17, of the case n = 4 was given in Ch. VIII.] 

 The treatment for n is similar to that for his first case n = 5. Then, by 

 addition, the sum of the 5 squares i, , si is 4a. First, find a square P 2 

 approximately equal to 4a/5, say 



z = 



5 ' 25z 2 ' m 2 -20a 



Since every number is a sum of 5 squares, set 



\-g 5 , 



2 n (Xi 



= =#;+, Si=gi+ai 



Thus ai = M-giN. Since Ss- = 4a, we get x= -2S0 t -a,-/Sa-. Thus, if 

 a = 21, the nearest square root of 20ais m = 21, whence z = 2, M = 41, N = W. 

 Since 4a = 1+9+25+49, l = (9+16)/25, the g's are 3/5, 4/5, 3, 5, 7, the 

 's are 35, 33, 11, -9, -29, and x = 1676/16785. 



To find 48 three numbers x, vx, v 2 x in geometrical progression such 

 that each increased by a given number n is a square. From +n = c 2 , 

 vx-\-n= (d+c) 2 , we get x, v. In the resulting value of v*x-\-n, put c 2 n = r 2 ; 

 then 



<Z 4 +4d 3 c+2d 2 (2c 2 +r 2 )+4r 2 dc+r 2 c 2 = D = (d 2 -2rd-rc) 2 



gives d. The desired numbers are r 2 , fr 2 n, (|r 2 ri) z /r 2 , where 

 r= (n s 2 )/(2s) makes r 2 +n= D =c 2 . 



Several 49 found four integers whose sum is a 2 and excess of the sum of any 

 three over the fourth is a square 6 2 , c 2 , <i 2 or e 2 . Hence 6 2 +c 2 +d 2 +e 2 = 2a 2 , 

 which determines a rationally if we take c = p a, d = q a. 



To find 490 two numbers (v z n and w 2 n) whose difference is a square 

 and such that if each and their sum be increased by the same number n 

 there result squares, we have to make v 2 w 2 and v 2 -\-w z n squares and 

 hence a certain quartic function a square. 



J. Win ward 50 found N integers whose sum is a square m 2 and sum of any 

 N 1 of them is a square. Take (2m n)n, (2m 2ri)(2n), (2m 3n)(3w), 

 , {2m (N l}n}(N \}n as the first 2V 1 numbers, and m 2 less their 

 sum as the Nth. Then m 2 exceeds the jth number (j<N) by (mjn) z . 

 Equating the excess of m 2 over the Nth number to (nr) 2 , we get m in terms 

 of n, r. 



Several 51 solved 2+a 2 = D, z/n-}-d 2 = D by known methods. 



To find 52 four integers whose differences are squares, let x = 2lmn, 

 y = l(m? n 2 ). Then five of the differences of u, u+x 2 , 



* 7 Mem. Acad. Sc. St. Petersbourg, 1, 1809, 271-282.* 



48 The Gentleman's Math. Companion, London, 2, No. 13, 1810, 264-5. 



49 The Gentleman's Diary, or Math. Repository, London, No. 71, 1811, 35, Quest. 963. For 



3 numbers, Gentleman's Math. Companion, 5, No. 29, 1826, 3624. 

 490 New Series of Math. Repository (ed., Leybourn), 3, 1814, I, 105-8. 



50 The Gentleman's Math. Companion, London, 5, No. 25, 1822, 141-2. 

 B1 Ladies' Diary, 1823, 35-36, Quest. 1390. 



62 The Gentleman's Math. Companion, London, 5, No. 26, 1823, 202-4. 



