Geometric/arithmetic progression of divisor, quotient, remainder

[Max A.]

On 1/14/07, Nick Hobson <nickh at qbyte.org> wrote:

> > In the second case, the number n=r*(r*m^3 + 1) gives perfect powers
> > quite rarely. In particular, since gcd(r,r*m^3 + 1)=1, both r and
> > r*m^3 + 1 must be a perfect power (of the same degree). This can never
> > give a cube, since if r is a cube, say, r=s^3, then
> > r*m^3 + 1 = (s*m)^3 + 1 cannot be a cube.
> >
>
> This is true if r*m^3 is an integer.

Yes, I did not consider the case when m is non-integer rational number.

[...]

> I haven't yet worked out a simple form for gcd(bu, u*a^3 + b).  I can see
> that any divisor d of bu and u*a^3 + b divides u(u*a^3 + b) - bu =
> u^2*a^3, which helps a little as gcd(a, b) = 1.

Let gcd(b,u)=d and u=dv, b=dc, and gcd(v,c)=1. Since gcd(a,b)=1 we
have also gcd(a,c)=1. Then
n = b*u*(u*a^3 + b) = d^3*v*c*(v*a^3+c), and it is easy to check that
gcd(v*c,v*a^3+c)=1.
Therefore, if n is a cube then each of v, c, and v*a^3+c is a cube.
But if v=w^3, c=e^3, and v*a^3+c=f^3, then (w*a)^3 + e^3 = f^3, which
is impossible.
Hence, there are no cubes in your sequence.

Max