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Notes on Explorations^409

E.30. The sequences ~~(4) and ~~(6) arise in the following context. The

number 5040 is not only 1 less than a perfect square, but differs from each

of the next three larger perfect squares by a perfect square:

5040 = 712 - l2 = 722 - 122 = 732 - 172.

The existence of other such numbers turns on making 8k2 + 1 and 3k2 + 1
both squares for a fixed value of k. Now 8k2 + 1 is square k = un(6)
for some n, and 3k2 + 1 is square k = u*(4) for some n. Hence the
question arises as to what numbers ~~(4) and ~~(6) have in common.
In the paper, M. Mignotte, Intersection des images de certaines suites
recurrentes lineaires, Theor. Comp. Sci. 7 (1978), 117-122 it is shown that
there are only finitely many such common numbers. Remarkably, in A.
Baker & H. Davenport, The equations 3x2 - 2 = y” and 8x2 - 7 = .r2,
Quart. J. Math. (2) 20 (1969), 129-137 a similar problem is solved using
deep results in diophantine approximation. See also E.J. Barbeau, Num-
bers differing from consecutive squares by squares, Canud. Math. Bull. 28
(1985), 337-342. See Notes on Explorations E.5 and E.29.
A similar problem involving sums instead of differences is E3080 found
in Amer. Math. Monthly 92 (1985), 215; 95 (1988), 141.
The paper by Carlitz and Thomas cited in the notes for Exploration E.7
indicates how the sequence {u,,} is tied in with the sequence involved in
the reciprocal substitution.
E.31. The polynomial has rational zeros if and only if its discriminant
5(n + 1)2 - 4 is a perfect square. Since 5(n + 1)2 - 4 has the same value
for n = m and n = -2 - m, it suffices to determine the situation for
nonnegative n. A little experimentation reveals that the discriminant is
square when n + 1 takes alternate values of the Fibonacci sequence, i.e.
when n + 1 is one of 1, 2, 5, 13, 34, 89,.... Indeed

ot2+t-2=r--2

t2 + 2t - 3 = (t - 1)(t + 3)
4t2 + 52 - 6 = (4t - 3)(t + 2)
12t2 + 13t - 14 = (3t - 2)(4t + 7)
33t2 + 34t - 35 = (11t - 7)(3t + 5)
88t2 + 89t - 90 = (8t - 5)( llt + 18).
Finding a complete set of n amounts to solving the diophantine equation
x2 - 5y2 = -4 (and then taking n = y - 1). A complete set of solutions is
given by the recursion (xi,yi) = (1,1) and

txn+l> Yn+l) = ((3% + 5~,,)/2, (xn + 3y,,)/2).