5 The Modular Group 201Since the second term on the right is divisible byp^2 ,wemusthavexv =εyuor
xu=−εyv. Evidentlyε=1inthefirstcaseandε=−1 in the second case. Since
(x,y)=(u,v)=1, it follows that eitherx=u,y=vorx=v,y=u.
The equationx^2 −dy^2 =1, wheredis a positive integer which is not a per-
fect square, is generally known asPell’s equation, following an erroneous attribution
of Euler. The problem of finding its integral solutions was issued as a challenge by
Fermat (1657). In the same year Brouncker and Wallis gave a method of solution which
is essentially the same as the solution by continued fractions. The first complete proof
that a nontrivial solution always exists was given by Lagrange (1768).
Unknown to them all, the problem had been considered centuries earlier by Hindu
mathematicians. Special cases of Pell’s equation were solved by Brahmagupta (628)
and a general method of solution, which was described by Bhascara II (1150), was
known to Jayadeva at least a century earlier. No proofs were given, but their method
is a modification of the solution by continued fractions and is often faster in practice.
Bhascara found the fundamental solution of the equationx^2 − 61 y^2 =1, namely
x= 1766319049 , y= 226153980 ,a remarkable achievement for the era.
5 The Modular Group
We recall that a complex numberwis said to beequivalentto a complex numberzif
there exist integersa,b,c,dwithad−bc=±1suchthat
w=(az+b)/(cz+d).Since we can write
w=(az+b)(cz ̄+d)/|cz+d|^2 ,the imaginary parts are related by
Iw=(ad−bc)Iz/|cz+d|^2.ConsequentlyIwandIzhave the same sign ifad−bc=1 and opposite signs if
ad−bc=−1. Since the mapz→−zinterchanges the upper and lower half-planes,
we may restrict attention toz’s in theupper half-planeH ={z∈C:Iz> 0 }and
tow’s which areproperly equivalentto them, i.e. withad−bc=1.
Amodular transformationis a mapf:H→Hof the form
f(z)=(az+b)/(cz+d),wherea,b,c,d∈Zandad−bc=1. Such a map is bijective and its inverse is again
a modular transformation:
f−^1 (z)=(dz−b)/(−cz+a).