### Arithmetic Progressions in Power Sequences

For a non-zero real number c, consider the infinite sequence

S(c)={1c, 2c, 3c . . . } .

What is the cardinality l(c) of the longest arithmetic progression, contained in S(c) ?

For instance, there are 3-terms arithmetic progressions, composed of perfect squares (like 1, 25, 49). On the other hand, Fermat proved that no four squares form an arithmetic progression; therefore, l(2)=3.

A two-minutes meditation shows that if c is the reciprocal of a positive integer (that is, c=1/n for a positive integer n), then S(c) contains an infinite arithmetic progression. A five-minutes meditation shows then that if l(c) is finite, then l(-c) is also finite and l(-c)=l(c); furthermore, if c is a reciprocal of a negative integer, then S(c) contains an arithmetic progression of any preassigned length. If S(c) contains no infinite progression and no progression of any preassigned length, then l(c) is finite.

It is a less trivial (but still not too complicated) task to prove that if c is not the reciprocal of a positive integer, then S(c) does not contain an infinite progression. This, however, leaves the following question open: do there exist any c, except for c=±1/n (with a positive integer n), such that S(c) contains progressions of any preassigned length?

Conjecture 1. If c is not a reciprocal of an integer number, then l(c) is finite.

Using a very strong result due to Darmon and Merel, I can actually determine l(c) for any rational c. Specifically, if c=p/q, where p and q are positive co-prime integers, then

• if p=1, then l(c) is infinite;
• if p=2, then l(c)=3;
• if p>2, then l(c)=2.
Of course, this establishes the conjecture above for all rational c. At the same time, for c irrational I can say virtually nothing about the value of l(c). I would not be surprised if the following is true.

Conjecture 2. If c is irrational, then S(c) can possibly contain at most one triple of elements in an arithmetic progression. In particular, l(c) is at most 3 for any irrational c.