Number Theory: An Introduction to Mathematics

4 Quadratic Diophantine Equations 195

But

− 1 /ξ 1 ′=ξ+a 0 =[2a 0 ,a 1 ,...,ah].

Comparing this with the previous expression, we see that (ii) holds.
Suppose, conversely, that (ii) holds. Thenξis irrational,a 0 >0 and henceξ>1.
Moreoverξ 1 =[a 1 ,...,ah] is a reduced quadratic irrational and

− 1 /ξ 1 ′=[ah,...,a 1 ]=[2a 0 ,a 1 ,...,ah]=a 0 +ξ.

Henceξ′=a 0 + 1 /ξ 1 ′=−ξandξ^2 =−ξξ′is rational.

4 Quadratic Diophantine Equations..............................

We are interested in finding all integersx,ysuch that

ax^2 +bxy+cy^2 +dx+ey+f= 0 , (6)

wherea,...,fare given integers. Writing (6) as a quadratic equation forx,

ax^2 +(by+d)x+cy^2 +ey+f= 0 ,

we see that if a solution exists for somey, then the discriminant

(by+d)^2 − 4 a(cy^2 +ey+f)

must be a perfect square. Thus

(b^2 − 4 ac)y^2 + 2 (bd− 2 ae)y+d^2 − 4 af=z^2

for some integerz. If we put

p:=b^2 − 4 ac, q:=bd− 2 ae, r:=d^2 − 4 af,

we have a quadratic equation fory,

py^2 + 2 qy+r−z^2 = 0 ,

whose discriminant must also be a perfect square. Thus

q^2 −p(r−z^2 )=w^2

for some integerw. Thus if (6) has a solution in integers, so also does the equation

w^2 −pz^2 =q^2 −pr.

Moreover, from all solutions in integers of the latter equation we may obtain, by
retracing our steps, all solutions in integers of the original equation (6).
Thus we now restrict our attention to finding all integersx,ysuch that
x^2 −dy^2 =m, (7)

wheredandmare given integers.
The equation (7) has the remarkable property, which was known to Brahmagupta
(628) and later rediscovered by Euler (1758), that if we have solutions for two values
m 1 ,m 2 ofm, then we can derive a solution for their productm 1 m 2. This follows from
the identity

(x^21 −dy 12 )(x 22 −dy^22 )=x^2 −dy^2 ,