carlosbr wrote: $15{th}$ Iberoamerican Olympiad Merida, VENEZUELA. [2000] Problem 4 From an infinite arithmetic progression $1,a_1,a_2,\dots$ of real numbers some terms are deleted, obtaining an infinite geometric progression $1,b_1,b_2,\dots$ whose ratio is $q$. Find all the possible values of $q$. $\text{\LaTeX}{}ed$ by Carlos Bravo - [carlosbr] http://www.kalva.demon.co.uk/ibero/isoln/isol004.html

The link 's died already.

It`s obvious that $ b_i=q^i$ Let $ d$ is the ratio of the arithmetic progression. Then $ q=1+n_1d$ ($ n_i$ is a natural number for all $ i$) $ \implies \frac{q-1}{d}=n_1$ And $ q^2=1+n_2d$ $ \implies \frac{(q-1)(q+1)}{d}=n_1$ Then $ q+1$ is a rational number and $ q$ is a rational number too. So we can let $ q=\frac{a}{b}$ where $ a$ and $ b$ are integers (relative primes). But it is true that $ q^v=1+n_vd$ $ \implies \frac{q^v-1}{d}$ is a natural number. Then: $ \frac {a^v-b^v}{b^v.d}$ is a natural number. But it is wrong because we can take $ v$ very large and it is impossible. Hence the possible values of $ q$ are all the naturals, doing $ d=1$(it obvious that if $ q$ is a negative integer don`t exist $ d$).

Just refreshing the link: http://web.archive.org/web/20040127230808/http://www.kalva.demon.co.uk/ibero/isoln/isol004.html

First note that $q$ could be any positive integer since if we take the arithmetic progression $1, 2, 3, 4...$ and leave the powers of a fixed random integer, it would follow that $q$ is such integer. It cant be a negative number because if it were, there would be an $n$ sufficiently large for which $a_m$ is negative if $m\geq n$ and this is impossible because even powers of $q$ would be negative. Let $d$ be the difference of the arithmetic progression, and let $r$ be the integer for which $b_1 =1+rd=q$. The numbers $q^2, q^3...$ are part of the arithmetic progression, \[q^2=1+2rd+r^2d^2=1+(2r+r ^2d)d \]\[q^3=1+(3r+3r^2d+r^3d^2)d\]\[q^4=1+(4r+6r^2d+4r^3d^2+r^4d^3)d\]\[...\]and this means that the quantities in parenthesis must be integers, which is equivalent to $r^{n+1} d^{n} \in \mathbb{N}$ by the binomial theorem.Since $r$ is an integer $d$ can't be irrational. Suppose it is rational but not an integer, WLOG we can assume $d=\frac{1}{x}, x \in \mathbb{N}, x>1$ and let $\gcd(x,r)=t$. From the relation abovementioned $(x')^nk=(r')^nr$ for some $k \in \mathbb{N}$ and $x=x't, r=r't$. Take a prime $p$ such that $p|x'$, the equation tells us that $p^n |r$, since $n$ is arbitrary this should hold no matter how big $n$ is, but this is impossible. Thus $\gcd (x,r)=x \rightarrow x|r$ resulting in $q \in \mathbb{N}$.