Quotients of Sums and Differences of Perfect Squares 5 



Every recurring continued fraction, whether pure or mixed, is 

 equal to some quadratic surd number. 

 We return to equation [13]. By [6] 



Po,n/pi.n = ( «o,<^i «« )>i 



since a^ and a„ are positive integers. Consequently 



/'0,n>A,«-l > or Pl,n-1 — Po,n=fO, 



and hence, 



A pure recurring continued fraction can not be equal to a pure 

 quadratic surd number. 



If «o = o, [3] gives us 



Po,n = P2,n, Po,n~l = p2,n-l, 



and [13] goes over into 



^ __ pLn~l—p2,n J\- ■\(pl.n-l—p2,ny-{-4Phnp2.n-l ^ [14] 



2A,w 2pi^„ 



Now by [3] and [4] 



Pl,n-l—p2,n = aiP2,n-l-\-Ps,n-l — («;z/«-1.2+A-2.2) 



hence in order that the rational constituent in [14] may vanish we 

 must have 



^1 A,«-l"f"A,«-l <^npn-l,2 — P>i-2,2=0, 



or 



p2,n-2—p%.n-\ [15] 



^1,=^« + 



p2n-\ 



Furthermore p2,n-\l p\n-2^={ cin-\. • • • , «2 ) ^ind p2,n-i/pz,n-i-= 

 (^2, ■ • •. a,i-\) hence /2,n-i is greater than either />2.»-2 or p^^„-i or 

 their difference pi,n-2— Pd.n~\> ( p2.n-2— P\n~\) I P-i.n-\ is a proper 

 fraction, and [15] is satisfied only when simultaneously 



ay=an and p2n-2 — P3,n-i—o. 



We now treat />2,«-2 — A.«-i precisely as we treated /i,«-i — p2,n 

 and find that p2,n-2 — Pz,n-\ can vanish only, if 



«2 = ^«-l ^rid Pz.n-Z — /'4,«-2=0, 



and this in turn leads to the general condition a;^+i=a„_^. 



359 



