Prove that for any positive integer $ m$ there exist an infinite number of pairs of integers $(x,y)$ such that $(\text{i})$ $x$ and $y$ are relatively prime; $(\text{ii})$ $x$ divides $y^2+m;$ $(\text{iii})$ $y$ divides $x^2+m.$
Problem
Source: Italy TST 2002
Tags: algebra, polynomial, Vieta, number theory, relatively prime, number theory unsolved
Rust
23.09.2008 19:35
Solve Pell's equation $ x^2+y^2+m=kxy$.
outback
23.09.2008 21:37
Rust, please show more details.
Rust
23.09.2008 21:45
Your conditions equavalent to $ xy|x^2+y^2+m$, therefore we can solve $ x^2+y^2+m=kxy$ and show, that for one k it had infinetely mane solutions.
outback
23.09.2008 21:52
Rust wrote: therefore we can solve $ x^2 + y^2 + m = kxy$ and show, that for one k it had infinetely mane solutions. I hate to say this, but can you show it?
nayel
23.09.2008 22:20
Fix $ k=m+2$. Then you can generate infinitely many solutions $ (x_n,y_n)$ using the following recursion: i. $ x_0=y_0=1$ ii. $ x_{n+1}=y_n$, $ y_{n+1}=(m+2)y_n-x_n$
Here I used Vieta Jumping for the equation $ x^2-(m+2)xy+y^2+m=0$. All we need to see is that $ (1,1)$ is a solution, and if $ (x,y)$ is a solution then so are $ ((m+2)y-x,y)$ and $ (y,x)$.
Rust
23.09.2008 22:27
$ x=1,y=m+1,k=m+2$ always solution, and $ D=4-k^2<0$, therefore for $ k=m+2$ we had infinetely mane solutions.
WakeUp
09.11.2010 22:03
See also http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=220390