322 HISTORY OF THE THEORY OF NUMBERS. [CHAP, ix 



Several writers 92 found nine integers in arithmetical progression whose 

 sum of squares is a square. 



A. Martin 93 noted that the sum of the squares of the nine numbers 

 x 4y, x 3y, - , x + 4y in arithmetical progression is a square if 

 9z 2 + 6(ty 2 = D. Take y = 3z, x 2 + 60s 2 = (x + zp/q) 2 ; hence x/z is 

 found rationally. 



Various writers 94 made SJz* Xf and I,x\ squares for n = 2, 3, 4, 5, 9. 



A. Boutin 95 noted values n = 4, 9, , 50 such that the sum of the 

 squares of n integers in arithmetical progression is a square. 



A. Martin 96 solved b\ + + &i = c} + + l by setting c n = a + b m 

 and finding b m rationally. 



T. Meyer 97 gave solutions of a 2 + b 2 + + n 2 + x 2 = z 2 . 



G. La Marca 98 proved that Za< = D if a\, -,a n are integers such that 

 i : a 2 = 3 : 4, a,- : a i+1 = 3:5 (i = 2, , rc 1). For, by a t = 3<?i, 

 02 = 4<?i, a 2 = 3g2, as = 5<? 2 , we have a 2 + 2 = (5<?i) 2 , 5<?i : a 3 = 3 : 4, 

 (5g0 2 + 3 = (5;z) 2 , etc., where z = 5<? 2 /4 is stated erroneously to be q z . 



Ed. Collignon" noted that x = 2ak(k + 1) is a solution of 



x 2 + (x - a) 2 + + (x - ka) 2 = (x + a) 2 + + (x + ka) 2 . 



E. N. Barisien 100 noted that a sum of p consecutive squares is not a 

 square for p < 20, except for p = 2, 11, without treating the case p = 13. 

 First, let p = In + 1 and denote the middle square by x 2 and the least 

 square by (x n) 2 . The sum of the squares is (2n + 1) {z 2 + n(n + l)/3}, 

 which is not a square for n ^ 4, n = 7, 8, 9. For n = 5, ll(x 2 + 10) is 

 to be a square, whence x = llh 1. Then x 2 + 10 = llm 2 , h = 21, 1 = 

 or 1 (mod 3). A table of 8 solutions includes 



(x, h, m) = (23, 2, 7), (43, 4, 13), (461, 42, 139), (859, 78, 259). 

 For p = 2n, let (x + n) 2 be the largest square. Their sum is 



N = 2nx(x + 1) + n(2n 2 + l)/3. 



For n = 1, N = 2x z -}- 2x -\- 1 =4^+1, where I 7 is a triangular number. 

 Thus T = 6, 210, 7158, , giving 



3 2 + 4 2 = 5 2 , 20 2 + 21 2 = 29 2 , 119 2 + 120 2 = 169 2 . 



The cases 1 < n ^ 9 are impossible. Cf. Lucas. 70 

 E. N. Barisien 101 gave the identity 



(a 2 + 6 2 + c 2 ) 3 = [a(& 2 + c 2 - a 2 )] 2 + [b(b 2 + a 2 - 3c 2 )] 2 

 + [c(a 2 + c 2 -- 36 2 )] 2 + (2a 2 6) 2 + (2a 2 c) 2 + (4a6c) 2 , 



92 Amer. Math. Monthly, 2, 1895, 129-30, 163. 

 83 Math. Quest. Educ. Times, 63, 1895, 111-2. 

 M L'interme'diaire des math., 1, 1897, 42-4. 



95 Ibid., 5, 1898, 75. 



96 Math. Magazine, 2, 1898, 212-3. 



97 Zeitechr. Math. Naturw. Unterricht, 36, 1905, 337-340. 



98 II Boll. Mat. Giornale Sc. Didat. (ed., Conti), 5, 1906, 152-5. 



99 Sphinx-Oedipe, 1906-7, 129. Case a = 1, Dostor. 75 



100 Sphinx-Oedipe, 1907-8, 121-6. Cf. Martin. 87 



101 Bull, de math. 616m., 15, 1909-10, 181. 



