Number Theory: An Introduction to Mathematics

4 Quadratic Diophantine Equations 199

of (10). Since

u=x 0 u 0 −dy 0 v 0 =x 0 u 0 −[(x 02 − 1 )(u^20 −m)]^1 /^2 > 0 ,

we must have

x 0 u 0 −dy 0 v 0 ≥u 0.

Hence

(x 0 − 1 )^2 u^20 ≥d^2 y 02 v^20 =(x 02 − 1 )(u^20 −m).

Thus

(x 0 − 1 )/(x 0 + 1 )≥ 1 −m/u^20 ,

which impliesu^20 ≤m(x 0 + 1 )/2 and hencedv^20 ≤m(x 0 − 1 )/2.
For the equation (11) we begin in the same way. Then from

(x 0 v 0 )^2 =(y 02 + 1 /d)(u^20 +m)>y 02 u^20

we obtainv=x 0 v 0 −y 0 u 0 >0 and hencex 0 v 0 −y 0 u 0 ≥v 0. Thus

d(x 0 − 1 )^2 v 02 ≥dy^20 u^20

and hence

(x 0 − 1 )^2 (u^20 +m)≥(x 02 − 1 )u^20.

The argument can now be completed in the same way as before.

The proof of Proposition 11 shows that if (10), or (11), has an integral solution,
then we obtain all solutions by finding thefinitely manysolutionsu,vwhich satisfy
the inequalities in the statement of Proposition 11 and composing them with all solu-
tions inCof (8).
The only solutionsx,yof (8) for whichx^2 ≤(x 0 + 1 )/2 are the trivial onesx=
±1,y=0. Hence any solution of (8) is inCor is obtained by reversing the signs of a
solution inC.
Ifu,vis a positive solution of (9) such thatu^2 ≤(x 0 − 1 )/2,dv^2 ≤(x 0 + 1 )/2,
thenx = u^2 +dv^2 ,y = 2 uv is a positive solution of (8) such thatx ≤ x 0.
Hence(x,y)=(x 0 ,y 0 )is the fundamental solution of (8) andu^2 =(x 0 − 1 )/2,
dv^2 =(x 0 + 1 )/2. Thus(u,v)is uniquely determined and is the fundamental solution
of (9). Hence, if (9) has a solution, any solution is obtained by composing the funda-
mental solution of (9) with an element ofCor by reversing the signs of such a solution.
A necessary condition for the solubility in integers of the equation (9) is thatd
may be represented as a sum of two squares. For the periodhof the continued fraction
expansionξ=

√

d=[a 0 ,a 1 ,...,ah] must be odd, sayh= 2 m+1. It follows from
Proposition 9 that

ξm+ 1 =[am,...,a 1 , 2 a 0 ,a 1 ...,am],