a^3 = amn + m + n

[Max]

Joshua Zucker wrote:
> Hi seqfans,
> someone asked me to prove that a^3 = amn + m + n had no solutions in 
> positive integers if a is prime.
> I haven't been able to do that yet.
> 
> But, seqfan that I am, I figured I could make the sequence of all a 
> which did solve this thing.
> 
> It turns out that a = x^3 works, with m = x^5 - x and n = x (of course m 
> and n are interchangeable; let's just call m the larger of the two).
> 
> And, besides those, if (a,m,n) works, then (m,?,a) also works.
> For instance, since (8,30,2) works, so does (30,112,8).
> And then since that works, so does (112,418,30)
> And thence (418,1560,112).
> 
> It appears that every solution -- at least up to a = 3000, which is as 
> far as I checked -- has one of these forms: either a = x^3, or the 
> solution is generated from a smaller one.
> 
> Can anyone help me prove that conjecture?

Rewriting the original equality as
a^3 - m = n (am + 1)
the problem becomes equivalent to findind all pairs (a,m) such that (am + 1) divides (a^3 - m).

Note that (am + 1) divides (a^3 - m) if and only
if (am + 1) divides
a^2*(am + 1) - m*(a^3 - m) = a^2 + m^2

This is a well-known problem. All such pairs (a,m) are consecutive elements of linear recurrent sequences
s(k+1) = s(1)^2 * s(k) - s(k-1)
for s(0)=0 and various s(1). For example,

s(1)=2: 0, 2, 8, 30, 112, 418, 1560, 5822, 21728, 81090, 302632, ...

s(1)=3: 0, 3, 27, 240, 2133, 18957, 168480, 1497363, 13307787, 118272720, 1051146693, ...

It is easy to see that s(2) is always a cube that proves your conjecture.

Max