70 PROCEEDINGS OF SECTION A. 
1680. 1681n? — 2. 1681. 1682nm + 1681. 1682m?2 
= 2 (492n — 493m) (2870n — 2871m) 
2. 2870262 — (2870 + 2871)%be + 2728717? 
= (1681b — 1682c) (98005 — 9801c) 
9799. 9800n? — 2. 9800. 9D80Inm + 9801. 9802m!* 
= 2 (2870n — 2871m) (16730n — 16731m) 
2. 1673026? — (16730 + 16731)%be + 2. 16731%c? 
= (98006 — 980Ic) (571216 — 57122c) 
57120. 57121n* — 2. 57121. 57122nm + 57122. 57123m!* 
= 2 (16730n — 16731m) (97512n — 97513m) 
2. 975122b? — (97512 + 97513)%be + 2. 97513%c? 
= (571216 — 57122c) (3329286 — 332929c) 
332927. 332928n? — 2.332928. 332929nm 
+ 332929. 332930 m* 
= 2(97512n — 97513m)(568344n — 568345m) 
2. 568344b* — (568344 + 568345)%e + 2. 5683457c% 
= (3329286 — 332929c)(19404496 — 1940450c) 
1940448. 1940449n* — 2. 1940449. 1940450nm 
+ 1940450. 1940451m? 
= 2 (568344n — 568345m)(3312554)n — 3312555m) 
2.3312554°b* — (3312554 + 3312555 )*be + 2.3312555%c* 
— (1940449d — 1940450c)(113097685 — 11309769c) 
ee degree m AS : 
_- Inrelation (1) for i and = Write w and a respectively ; it 
. then reduces to 
. 74 (2a — 1) — 2ua + 7 ae) en 0 ae Se (11) 
From (ii) we obtain 
wae ae moi {i+ i= x) ve p> 
~and hence 2a must be a perfect square. 
The values of a found from the above series are 
2 9 50 289 1682 9801 
1’ 8’ 49° 288 1681’ 9800” 
Hence a may be represented by one or other of the 
Aes ay a 
5 OR ae 
forms and we get the following , 
table :— 
