Unknown

Notes on Explorations

2y2 - %2 = 1.

The positive solutions to equation (1) are given by

where

(Xk,Yk) = (l,l), (3,5), (11,19), (41,71),...

xk+l = 2xk + yk = 4tk - xk-1

Yk+l = 3xk + 2yk = 4yk - yk-1.

The positive solutions to equation (a) are given by

(Yj, 4 = (1, 11, (5,7), G-%41), * * *

where

yj+l = 3yj + 2Zj = 6yj - yj-1

Zj+l = 4yj + 3%j = 6%j - %j-1.

397

(2)

Putting these together, we arrive at two solutions (z,y, z) = (l,l, l),

(3,5,7). Both these correspond to quadratics which are identically squares

of linear polynomials. The existence of a nontrivial quadratic turns on

whether the two “y-sequences”

1, 5, 19, 71, 265, 989, 3691,...

1, 5, 29, 169, 985, 5741,...
have a third integer in common. This is a difficult question to deal with.
For research into this type of problem, consult
A. Baker & H. Davenport, The equations 3x2 - 2 = y2 and 8x2 - 7 = z2
Quart. J. Math. (2) 20 (1969), 129-137.
K. Kubota, On a conjecture of Morgan Ward, I
Acta Arith. 33 (1977), 11-48.
R. Loxton, Linear recurrences of order two
J. Austrul. Math. Sot. 7 (1967) 108-114.
M. Mignotte, Intersection des images de certaines suites recurrentes lineaires
Theor. Camp. Sci. 7 (1978), 117-122.
P. Kanagasabapathy & Tharmambikai Ponnudurai, The simultaneous dio-
phantine equations y2 - 3x2 = -2 and .z2 - 8x2 = -7
Quart. J. Math. (2) 26 (1975), 275-278.
Quadratics taking four successive square values x2, y2, z2, w2 when the
variable takes the respective values 0, 1,2,3 are easy to find. The condition
is that
x2 + 3%2 = w2 + 3y?