The Pell equation, by Edward Everett Whitford.

THE PELL EQUATION

11

EB,
DB = DZ
AD2 + DZ2 = AE2 + EZ2 = 2AC2 + 2CD2.
If
CD = q and BD = p,
AD = 2q + p = p, AC = q+ p = l,
then
2q2 - p2 = (2q2 - p2).
This enables us to derive from one integral solution,
p, q, of one of the two equations
2 - 2y2 = l 1,
a solution, pi, q1, for the other, always in larger integers.
We might start from 1, 1, or from the one with which the
Pythagoreans were already familiar, 7, 5.
As Proclus1 pointed out, the Pythagoreans proceeded
as follows:2 On AB (Fig. 2) construct a square and draw
its diagonal BE. On the prolongation of AB lay off
K
E \
A B C D G H
FIG. 2.
BC = AB and CD = BE. Then according to the Pythagorean theorem
CD2 = 2AB2,
and by the theorem just referred to
AD2 + CD2 = 2AB2 + 2BD2
1 Proclus, loc. cit.
2 Hultsch, loc. cit.