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Let $x,y \in \mathbb{R}$ such that the set $\{\cos{(n\pi x)} + \cos{(n\pi y)} | n \in \mathbb{N} \}$ is finite. Show that $x$ and $y$ are rational.

I have been trying to consider a graph of this set and maybe an argument by contradiction....

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2 Answers

Lemma. If $\alpha\in\mathbb R$, $m\in\mathbb N$, there exist $a,b\in\mathbb Z$ with $1\le a\le m$ and $|a\alpha-b|<\frac 1m$.

Proof: Partition the interval $[0,1)$ into $m$ subintervals $\left[\frac km,\frac{k+1}m\right)$, $0\le k<m$. By the pigeon-hole principle, there exist $0\le i<j\le m$ such that $i\alpha-\lfloor i\alpha\rfloor$ and $j\alpha-\lfloor j\alpha\rfloor$ are in the same subinterval, hence with $a=j-i$ and $b=\lfloor j\alpha\rfloor-\lfloor i\alpha\rfloor$, the result follows.$_\square$

Corollary. If $\alpha,\beta\in\mathbb R$, $\epsilon>0$, there exist $r,s,n \in \mathbb Z$ such that $n\ge 1$, $|n\alpha-r|<\epsilon$ and $|n\beta-s|<\epsilon$.

Proof: Select $N>\frac1\epsilon$. By the lemma, there exist integers $a,b$ such that $1\le a\le N^2$ and $|a\cdot \alpha-b|<\frac 1{N^2}$. Again by the lemma, there exist integers $c,d$ such $1\le d\le N$ and $$|d\cdot a\beta-c|<\frac1N<\epsilon.$$ But we also have $$|da\alpha-db|<\frac d{N^2}\le \frac1N<\epsilon.$$ Thus the result follows with $n=da$, $r=db$, $s=c$. $_\square$


Now for the main event:

By continuity of $\cos$, for any $\epsilon>0$ there exists $\delta>0$ such that $\cos t>1-\frac\epsilon2$ for all $t$ with $|t|<\delta$. We find $r,s,n$ as in the corollary such that $$\left\vert n\cdot\frac x{2}-r\right\vert< \frac{\delta}{2\pi}\qquad\text{and}\qquad\left\vert n\cdot\frac y{2}-s\right\vert<\frac{\delta}{2\pi}.$$ It follows that $n\pi x$ and $n\pi y$ differ from $2\pi r$ and $2\pi s$ by less than $\delta$, hence $$\cos (n\pi x)+ \cos(n\pi y)>2-\epsilon.$$

Now assume that $S:=\{\cos(n\pi x)+\cos(n\pi y)\mid n\in\mathbb N\}$ is finite. Then $\max S<2$ would lead to a contradiction if we let $\epsilon = 2-\max S$. Therefore $\max S=2$, i.e. there exists $n\in\mathbb N$ with $$\cos (n\pi x)+ \cos(n\pi y)=2,$$ hence $\cos(n\pi x)=\cos(n\pi y)=1$ and $nx,ny$ are even integers, hence $x,y\in\mathbb Q$. $_\square$

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If $x$ is irrational, the values $e^{in\pi x}$ are dense on the unit circle. Similarly for $y$.

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I am sorry, but I don't see how this answers the question. When considering this, we have to consider sine as well, yes? –  Math2012pc Nov 26 '12 at 0:58
    
If those complex numbers are dense on the unit circle, then their real parts, which are the cosines, are dense in the interval $[-1,1]$. No need to consider the sines. –  Gerry Myerson Nov 26 '12 at 4:35
    
Ah, I see. But in order to show that the complex numbers are indeed dense on $S^1$, do we need to use a topological argument or can we argue with number theory? –  Math2012pc Nov 26 '12 at 5:08
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The sum of two dense sequences need not be dense! –  Hagen von Eitzen Nov 26 '12 at 7:25
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