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398 Notes on Explorations

Taking note of the special case k = 3 of the identity

(p2 + kq2)(r2 + ks2) = (pr + kqs)2 + k(ps - qr)2

= (pr - kqs)2 + k(ps + qr)2,

we can take x = pr - 3qs, y = ps - qr, z = ps + qr, w = pr + 3qs. For
example, (p,q, r, s) = (4,1,2,1) leads to (x, y, z, w) = (5,2,6,11) and the
quadratic (1/2)(53t2 - 95t + 50).
But more can be said. D. Allison gives the example

-420t2 + 2100t + 2809

which is square for integers between -1 and 6 inclusive, and the example

-4980t2 + 321Oot + 2809

which takes distinct square values for integers between 0 and 6 inclusive.
Consult D. Allison, On square values of quadratics, Math. Proc. Camb.
Phil. Sot. 99 (1986), 381-383; and Duncan A. Buell, Integer squares with
constant second differences, Mathematics of Compulalion 49 (1987), 635-
644.
The “1986” problem is due to Andy Liu of the University of Alberta. He
bases a solution on the observation that

(t - l)(t - 9)(t - 8)(1 - 6) = (t2 - 12t)2 + 53t2 - 606t + 432.

The polynomial (t2 - 12t)2 - (t - l)(t - 9)(t - 8)(t - 6) takes a negative
value for t = 1986 and so provides a suitable example. P. Reiss of Winnipeg
considers the polynomial

f(t) = k(l - t)(t - 9) + r2

where k and r are to be chosen to make f(8) and f(6) squares, say u2 and
v2 respectively. We need

15(u2 - r”) = 7(v2 - r”).

A trial of u = r + 2 and v = r + 4 leads to r = 13 and k = 8. Thus
f(t) = -8t2 + 80t + 97 works. Another polynomial which works is 2t(9 - t).
Is it possible for a polynomial (not necessarily a quadratic) over Z to
assume a square value at every integer, and yet itself not be the square of
another polynomial? The answer is no, and we have the following general
result due to W.H.J. Fuchs:

Theorem. Suppose that f and g are two polynomials and that there is
an integer m such that, for each integer n > m, there is a number k such
that f(n) = g(k). Then there is a polynomial h such that f = g o h. If f