Given a positive integer n which is not a cube, define a = n1/3, b = 1/(a - [a]), c = 1/(b - [b]). Show that there are infinitely many such n for which we can find integers r, s, t, not all zero, such that ra + sb + tc = 0.
Solution
The basic idea is to express b and c as quadratics in a with rational coefficients. We can then choose r, s, t so that ra + sb + tc = 0. Put m = [a], then n-m3 = a3-m3 = (a-m)(a2+am+m2). That gives us a way to write b = 1/(a-m) as a quadratic. Inverting 1/(b-[b]) is trickier.
Put k = n-m3, so we have b = (a2+am+m2)/k. Obviously a - [a] < 1, so b > 1. We simplify things by taking b < 2, so that [b] = 1. We have a < m+1, so (a2+am+m2) < 3m2 + 3m + 1. So for suitable n (and hence k) we should be able to get 3m2 + 3m + 1 < 2k (*).
Then c = 1/(b-1) = k/(a2+am+m2-k). Suppose we factorise the denominator as (a-x)(a-y). Then 1/(a-x) = (a2+ax+x2)/(a3-x3). Now x+y = -m, xy = m2-k, so x3+y3 = (x+y)(x2-xy+y2) = (x+y)( (x+y)2 - 3xy) = -m(m2 - 3(m2-k) = m(2m2-3k). So if we put h = (a3-x3)(a3-y3), then h = n2 - n(x3+y3) + x3y3 = n2 - nm(2m2-3k) + (m2-k)3 is an integer. We have c = (k/h)(a2+ax+x2)(a2+ay+y2) = (k/h)(a4 + (x+y)a3 + (x2+xy+y2)a2 + xy(x+y)a + x2y2) = (k/h)(a4 - ma3 + (m2-m2+k)a2 - m(m2-k)a + (m2-k)2) = (k/h)(ka2 + (m(k-m2)+n)a + (m2-k)2 - mn), which is the required quadratic expression.
Now consider ra + sb + tc. To make the coefficient of a2 zero we need s/k + tk2/h = 0. The coefficient of a0 is then sm2/k + (tk/h)(m2-k)2 - mn) = (tk/h)( m2-k)2 - mn - km2). It is no good taking t = 0, so we have to get m2-k)2 - mn - km2 = 0 and hence 0 = m4 - 2km2 + k2 - mn - km2 = m(m3-n) - 3km2 + k2 = - mk - 3km2 + k2 = k(k - 3m2 - m). This suggests we should take k = 3m2 + m and hence n = m3 + 3m2 + m. That certainly gives [a] = m (since (m+1)3 = m3 + 3m2 + 3m + 1 > n) and also [b] < 2, since 3m2+3m+1 < 6m2+2m for m ≥ 1, which is what we needed at (*) above. There are obviously infinitely many n of this type (just take m = 1, 2, 3, ... ).
So if we take s = k3, t = -h, then s and t are integers and sb + tc = (mk2 - k((m2-k)2 - mn) )a, which is an integral multiple of a, as required.
© John Scholes
jscholes@kalva.demon.co.uk
9 March 2004
Last corrected/updated 9 Mar