Problem:
(a) Determine all nonnegative integers $r$ such that it is possible for an infinite arithmetic sequence to contain exactly $r$ terms that are integers. Prove your answer.
(b) Determine all nonnegative integers $r$ such that it is possible for an infinite geometric sequence to contain exactly $r$ terms that are integers. Prove your answer.
My Solution for $a)$:
$a)$ First, we can analyze the case when $r \geq 2$. In this case, there will be at least $2$ terms in the arithmetic sequence which are integers; $a+md$ and $a+nd$. WLOG, $m>n$. These terms are both integers, so $(m-n)d$ is also an integer. Adding $(m – n)d$ to $a + md$ gives $a + (2m – n)d$, which is also a member of the arithmetic sequence and an integer, so this process can continue, giving the arithmetic sequence an infinite number of terms. Therefore, $r$ can only be $r=1$.
I am currently stuck on $b)$, but I know that we could do a similar approach, but how would I do it?
Best Answer
For part (b):
Take a positive integer $m$.
Construct a GP with first term $m^{r-1}$ and common ratio $\frac 1m$. The GP is
$$\overbrace{ m^{r-1}, m^{r-2},m^{r-3},\cdots\cdots,m^2, m^1, 1}^{r \text{ integer terms}}, \overbrace{\frac 1m, \frac 1{m^2},\cdots\cdots\cdots}^{\text{non-integer terms}}$$