Does finite sequence mean that $ A_{x,y} $ take on only a finite no. of values???

From the equality \[\left( \cos n\pi x -\cos n\pi y \right) ^2=2+\cos2n\pi x +\cos2n\pi y -\left( \cos n\pi x +\cos n\pi y \right)^2 \]we deduce that the set \[B_{x,y}=\{\cos n\pi x -\cos n\pi y |n\in \mathbb{N}\} \]is also finite. Since \[\cos n\pi x =\frac{(\cos n\pi x +\cos n\pi y )+(\cos n\pi x -\cos n\pi y )}{2} \]we obtain that the set \[A=\{\cos n\pi x |n \in \mathbb{N}\} \]is finite as well. It follows that there exist $n,m \in \mathbb{N},n \ne m$ such that $\cos n\pi x=\cos m\pi x$, hence $x=\frac {2k}{n\pm m}$, for some integer $k$. Similarly, $y \in \mathbb{Q}$.

enescu where did you get that $\displaystyle B_{x,y}=\{\cos n\pi x -\cos n\pi y |n\in \mathbb{N}\}$ is finite? i really want to learn from this solution can u explain

Look at the right hand side of the equality \[\left( \cos n\pi x -\cos n\pi y \right) ^2=2+\cos2n\pi x +\cos2n\pi y -\left( \cos n\pi x +\cos n\pi y \right)^2 \]and observe that it only takes finitely many values, according to the hypothesis (because $\cos2n\pi x +\cos2n\pi y \in A_{x,y}$ as well)

thanks a lot really nice solution...

ehsan2004 wrote: Let $x,y\in \mathbb{R}$. Show that if $A_{x,y}={\{\cos n\pi x+\cos n\pi y \mid n\in \mathbb{N}\}}$ be the finite sequence then $x,y \in \mathbb{Q}$. I heard Vasile Pop is the author of this problem ? Is it true ? Are there some others articles related to this problem ?

Moubinool wrote: I heard Vasile Pop is the author of this problem ? Is it true ? Yes. Moubinool wrote: Are there some others articles related to this problem ? None to my knowledge.

Well, Vasile Pop wrote an article in Gazeta Matematica (I think in 1999 or later) about some of his problems given in national olympiads and among them this one was present too. He even generalized these problems. Unfortunately, I'm in Paris now and I don't have Gazeta Matematica with me.

same as the official solution i find in a book.

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