Show that the cube roots of three distinct primes cannot be terms in an arithmetic progression (whether consecutive or not).
Solution
Suppose the primes are p, q, r so that q1/3 = p1/3 + md, r1/3 = p1/3 + nd, where m and n (but not necessarily d) are integers. Then nq1/3 - mr1/3 = (n - m)p1/3. Cubing: n3q - 3n2mq2/3r1/3 + 3nm2q1/3r2/3 - m3r = (n-m)3p, or q1/3r1/3(mr1/3 - nq1/3) = ( (n-m)3p + m3r - n3q)/(3mn). But mr1/3 - nq1/3 = (m - n)p1/3, so we have (pqr)1/3 = ( (n-m)3p + m3r - n3q)/(3mn(m-n) ) (*).
It is now clear that we do not need p, q, r prime, just that pqr is not a cube, for then by the usual argument it must be irrational so that (*) is impossible.
© John Scholes
jscholes@kalva.demon.co.uk
19 Aug 2002