14 Robert E. Morits - 



apparently a quadratic form, but in essence linear. For since .r 

 and y are to be integers, x^ — y will be integral, and putting this 

 equal to 2 our equation becomes 



and it remains only to solve the congruence 



2A^ = —Ps,si^0^p2.2)- [27] 



If this congruence has a root Xo, which is always the case ex- 

 cept when 2/J3 and p^^„ possess a divisor which is prime to p^^^ 

 then 



pix.ai, . . . ,a2x) 



y z=:^ — : — ~ 



p{.ai ai) 



will represent an integer for every value 



xx= Xo-\- A./>2,2 



where X is any integers which will render x>, positive. 2>. will then 

 be given by 



Z. = -^ = -2XA,3-?^mA3. 



A.2 



and j^^ by 



The results just obtained are of importance in the theory of 

 recurring continued fractions and may be used to construct with 

 comparative ease a table much more extensive than the famous 

 Canon Pellianum. For example, the integral values x which 

 render l 



p{x,i,i,2,i,i,x) 

 /(i, 1, 2, 1,1) 



integral are given by the congruence 



r ^ 



2/>(i,2,i,i).r = — ;!)(i,2,i) mod p{i, 1,2, 1,1) 



368 



