Prove that the sequence $a_n = 3^n- 2^n$ contains no three numbers in geometric progression.
Problem
Source: Romania TST for IMO 1994
Tags: number theory
Math1331Math
06.09.2017 03:59
Very easy for olympiad title quite appropriate just compare $a_{n}a_{n+2}$ and $(a_{n+1})^2$ which gives immediate contradiction
TheMathateer
06.09.2017 04:05
@Medjl, why are you posting so many Romania problems? If they are "easy", why not try them yourself? Also, shouldn't there be an answer key online?
Math1331Math
06.09.2017 04:24
TheMathateer wrote: @Medjl, why are you posting so many Romania problems? If they are "easy", why not try them yourself? Also, shouldn't there be an answer key online? I believe it is for contest archives
mathwiz0803
06.09.2017 04:40
Math1331Math wrote: Very easy for olympiad title quite appropriate just compare $a_{n}a_{n+2}$ and $(a_{n+1})^2$ which gives immediate contradiction I don't think they have to be consecutive.
ThE-dArK-lOrD
06.09.2017 09:29
Zsigmondy's overkills this problem.
Medjl
06.09.2017 16:42
TheMathateer wrote: @Medjl, why are you posting so many Romania problems? If they are "easy", why not try them yourself? Also, shouldn't there be an answer key online? i solved most of them but for your information there aren't in the contest collection,of course some of them have been posted but i don't have time to seach for every one separatly, i posted both BMO tst and IMO TST. by the way you can try some of the Romania problems that i posted which are unsolved
Afo
12.10.2020 10:57
Solution. Assume the contrary. Then, for $x<y<z$, we have $$(3^y-2^y)^2 = (3^x-2^x)(3^z-2^z).$$By Zsigmondy's theorem, there is a prime $p|3^z-2^z$ such that $p \nmid 3^y-2^y$. A contradiction $\blacksquare$
PNT
11.03.2023 00:30
Is there a solution without Zsigmondy?
TerrificChameleon
11.03.2023 00:53
PNT wrote: Is there a solution without Zsigmondy? Heres mine,
Suppose for the sake of contradiction that there exist distinct positive integers $i,j,k$ such that $a_i, a_j,$ and $a_k$ are in geometric progression. That is, there exists a positive integer $r$ such that $a_j = ra_i$ and $a_k = ra_j$. Then, we have $$3^j - 2^j = r(3^i - 2^i) \quad\text{and}\quad 3^k - 2^k = r(3^j - 2^j).$$Subtracting the first equation from the second, we get $$3^k - 2^k - r(3^i - 2^i) = r(3^j - 2^j) - r(3^i - 2^i) = r(2^i)(3^{j-i}-1).$$Since $i,j,k$ are distinct positive integers, we have $j-i\geq 1$ and $k-i\geq 2$. Note that $3^{j-i}-1$ is odd and $2^i$ is even, so $r(2^i)(3^{j-i}-1)$ is divisible by $2^{i+1}$ but not by $2^{i+2}$.
Now, we have $2^k > 2^{k-1} + \cdots + 2^0 = 2^k-1$. Therefore, $2^k-1$ cannot be written as a sum of fewer than $k$ positive powers of $2$. In particular, $r(2^i)(3^{j-i}-1)$ cannot be written as a sum of fewer than $k$ positive powers of $2$, since $j-i\geq 1$ and $k-i\geq 2$. This means that $|3^k-2^k-r(3^i-2^i)| \geq 2^{i+1} > 1$, contradicting the assumption that $a_i,a_j,a_k$ are in geometric progression.
Therefore, the sequence $a_n = 3^n - 2^n$ contains no three numbers in geometric progression.
hectorleo123
18.05.2023 03:35
Medjl wrote: Prove that the sequence $a_n = 3^n- 2^n$ contains no three numbers in geometric progression. Suppose otherwise: First it is easy to see that the sequence is strictly increasing Let $a_n,a_m,a_k (m<n<k)$ be the numbers that are in the geometric progression Let r be the reason $r\in \mathbb{Q}$ $r=\frac{c}{d}$ By Zsigmondy's Theorem: $\exists$ prime $p/p|3^m-2^m, p\nmid3^n-2^n$ $\Rightarrow p|c$ By Zsigmondy's Theorem: $\exists$ prime $q\neq p / q|3^k-2^k, q\nmid 3^m-2^m$ $\Rightarrow q|c$ $\Rightarrow pq|c$ $a_nr=a_m$ $\Rightarrow q|3^m-2^m (\Rightarrow\Leftarrow)_\blacksquare$