# Possibilities for the number of integers in an infinite arithmetic or geometric sequence [closed]

(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.

## closed as off-topic by Lee Mosher, lulu, 6005, Chill2Macht, JMPAug 12 '16 at 2:49

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• Any thoughts? For a) can you, say, come up with an arithmetic progression with no integers? How about exactly one? exactly two? – lulu Aug 11 '16 at 23:57

For (b), consider $2^{r-n}$ for which only the first $r$ terms are integers.

(a) There are two variants:

1. r=0. This happens, for example, with a series $\frac{1}{2} + \sqrt{2}d$

2. r=1. Look at the series $0 + \sqrt{2}d$ In fact, suppose that $a$ and $a+sd$ are integers. Than all the numbers of the type $a + ksd$, where $k$ is integer, are also integers, so $r=\infty$

(b)

1. Look at the series $\sqrt{2}\exp^d$. It doesn't contain any integers.

2. Now look at the series $\exp^d$. It contains only one integer.

3. In fact the values $r$ can be whatever you want. Let $a = q^{r-1}$, $e = \frac{p}{q}$, $GCD(p,q)=1$. Then the only integers in the series $ae^d$ would be:

$$a = q^{r-1}$$

$$ae = \frac{p}{q^{r-2}}$$ $$...$$ $$ae^{r-1} = p$$

Hint For $(a)$ show that if an arithmetic progression contains two integers, the ratio must be a rational number.

Hint 2 If an arithmetic progression has ratio $r=\frac{k}{n}$ and a term $a_l \in \mathbb Z$ then you can make $$a_{l+m}=a_l + m \frac{k}{n}$$ an integer by setting $m$ to ......