Consider the sequence of rational numbers defined by $x_1=\frac{4}{3}$, and $x_{n+1}=\frac{x_n^2}{x_n^2-x_n+1}$. Show that the nu,erator of the lowest term expression of each sum $x_1+x_2+...+x_k$ is a perfect square.
Problem
Source: Romania 2017 IMO TST 1, problem 3
Tags: Sequence, number theory
tenplusten
18.03.2018 21:08
İ wonder what is original source Beacuse it was also Azerbaijan IMO TST 2017 .
GGPiku
18.03.2018 21:09
tenplusten wrote: İ wonder what is original source Beacuse it was also Azerbaijan IMO TST 2017 . Albania, on EGMO 2016 shl.
Anar24
19.03.2018 09:53
My team leader once told me that this problem initially appeared in Iran IMO TST(presumably in 1999 or 2000)
GGPiku
23.03.2018 22:03
Observe that the reccurence is equivalent with $x_n=\frac{1}{x_{n+1}-1}-\frac{1}{x_{n}-1}$
The sum will be $\sum \frac{1}{x_{k+1}-1}-\frac{1}{x_{k}-1}=\frac{1}{x_{k+1}-1}-3=\frac{(x_n-2)^2}{x_n-1}$. Denoting $x_n-1=\frac{a}{b}$, the sum will be $\frac{(a-b)^2}{ab}$. The conclusion follows.
silouan
26.05.2018 16:23
The same sequence appeared also here: https://artofproblemsolving.com/community/c6h53266p334356
test20
26.05.2018 16:34
The same sequence appeared also here: https://artofproblemsolving.com/community/c6h1641805