Sequences $x_1,x_2,\dots,$ and $y_1,y_2,\dots,$ are defined with $x_1=\dfrac{1}{8}$, $y_1=\dfrac{1}{10}$ and $x_{n+1}=x_n+x_n^2$, $y_{n+1}=y_n+y_n^2$. Prove that $x_m\neq y_n$ for all $m,n\in\mathbb{Z}^{+}$. Proposed by A. Golovanov
Problem
Source: St. Petersburg MO 2000, 10th grade, P1
Tags: algebra, sequances
StarLex1
22.04.2023 10:46
assume it exists for some $(m,n)$
$x_m=y_n$
thus
$x_{m-1}+x_{m-1}^2 = y_{n-1}+y_{n-1}^2$
forcing $x_{m-1} = y_{n-1}$
since $(x_k,y_k)>0$
repeat the process until one of them is 1
if $m>n$
$x_{m-(n-1)} = y_1$
$\frac{1}{10} = x_{m-{n-1}}$
however $\frac{1}{8}=x_1<x_2<.....<x_n$
if $m<n$
$x_1 = y_{n-{m-1}}$
$\frac{1}{8} = y_{n-m+1}$
let $n =m+k$
$\frac{1}{8} = y_{k+1}$
thus $y_{k}$ irrational however from a slight check we get all y_k is rational
if $m=n$ it is clearly impossible
RagvaloD
22.04.2023 14:54
Just note that denominator of $y_i$ is always divided by $5$ and denominator of $x_i$ is degree of $2$
grupyorum
22.04.2023 21:07
My sol is essentially the same as #2, but somewhat cleaner. Call $(m,n)$ good if $x_m=y_n$. Assume the contrary and take a good pair $(m,n)$ with the smallest coordinate sum $m+n$. Clearly $x_k>y_k$ for all $k$, so $n>m\ge 1$ must hold. Now I show that in fact $m=1$. Indeed if $m\ge 2$, then $x_{m-1}^2+x_{m-1} = y_{n-1}^2+y_{n-1}$ yields $x_{m-1}=y_{n-1}$ (as $x_k,y_k>0$ throughout). So, $(m-1,n-1)$ is also good, contradicting with minimality of $(m,n)$. Now we must have $y_n^2+y_n = x_1=1/8$, yielding $(2y_n+1)^2 = 3/2$. This is however a clear contradiction as $y_n\in\mathbb{Q}$.