Here's an idea: We can prove easily by induction, that for each integer $ k$ there exists a polynomial $ R_{k}$ so that $ R_{k} (cos x) = cos (kx)$. Suppose $ x = cos \theta$ is irrational. Then $ R_{k} (x) = - p$ and $ R_{k + 1} (x) = - q$ are both rational. Thus $ R_{k} (x) + p$ and $ R_{k + 1} (x) + q$ are polynomials with rational coefficients, which have zeros at $ x$. Consider the monic polynomial with rational coefficients $ Q$ of minimal degree such that $ S(x) = 0$. Using the well-known Lemma:[ if for an algebraic number $ x$ the rational polynomial of minimal degree which has $ x$ as a 0 is $ Q(x)$, then if $ R(x) = 0$ for some rational polynomial $ R(x)$, we have $ Q(x)$ is a factor of $ R(x)$.], we have $ Q(x)$ is a factor of both $ R_{k} (x) + p$ and $ R_{k + 1} (x) + q$, so $ R(x)$ is a factor of $ (R_{k} (x) + p, R_{k + 1} (x) + q)$. Since $ x$ is irrational, the degree of $ Q(x)$ is at least $ 2$. Our goal is to prove that the degree of $ (R_{k} (x) + p, R_{k + 1} (x) + q)$, and hence $ Q(x)$ is $ 2$. To do that, it may be possible to apply Euclidean algorithms to those 2 polynomials. Hopefully, there will be a special relationship between the coefficients of the polynomials that we obtain, which will make this easier to do. I think we can prove, using this method that the degree of $ (R_{k} (x) + p, R_{k + 1} (x) + q)$ is at most $ 2$, and if it is $ 2$ then $ p$ and $ q$ must take specific values (depending on the value of $ k (mod 6)$). If $ k \equiv 1 (mod 3)$, using this method I think we should get a contradictory relationships between $ p,q$, if the degree of the LCD polynomial is 2, that means $ k \equiv 1 (mod 3)$ is not possible. Once, for a certain value of $ k$, we find out what $ p$ and $ q$ are (one should be $ 0.5$ or $ - 0.5$, the other should be $ 0$), the problem will be finished. ---- Here's how we can kill $ k = 3$ using this method: Since $ P_{3} (x) = 4x^3 - 3x$ and $ P_{4} (x) = 8x^4 - 8x^2 + 1$, let $ 4x^3 - 3x + a = 0, 8x^4 - 8x^2 + b = 0$ for rational $ a,b$. Then if $ Q(x)$ is minimal integer polynomial of $ x$, $ Q(x)$ divides $ 4x^3 - 3x + a$ and $ 8x^4 - 8x^2 + b$, $ Q(x)$ has degree at least $ 2$. We have: $ (4x^3 - 3x + a, 8x^4 - 8x^2 + b) = (4x^3 - 3x + a, 2x^2 + 2ax - b) = (2x^2 + 2ax - b, 4ax^2 - (2b - 3) x - a)$. Here we have used typical Euclidean algorithm: using fact that $ 8x^4 - 8x^2 + b = (4x^3 - 3x + a)2x - (2x^2 + 2ax - b)$, etc. Since the GCD of those 2 polynomials, degree $ 2$, is a polynomial of degree $ 2$, either the $ 2$-nd one is $ 0$ - giving us $ b = 3/2, a = 0$ from where we get that $ cos (3 \theta) = 0$, $ cos (4 \theta) = - 0.5$ and $ \theta = \frac {\pi}{6}$ as required: OR the second quadratic is a rational multiple of the first. In this case, a routine calculation gives a contradic tion.

N.T.TUAN, excuse me, where did you find USA TST 2007? :

cheese: they've been posted here: http://www.mathlinks.ro/viewtopic.php?t=177771

Hm.. in last year, I saw this problem and tried to solve this problem by same method as yours.... I checked it for $ k=3,4,5$ but I failed to prove for all cases. in my case, It isn't enough to prove it, but it needs more facts such as $ |x|\le 1$...., please post details.. I want to see a solution of this problem.

I don't know why I couldn't think about this...... Let $ T_{n}$ be a integer coefficient polynomial such that $ T_{n}(\cos \theta) = \cos n\theta$. It is easy to prove that $ T_{n + 1}(x) + T_{n - 1}(x) = 2xT_{n}(x)$, Therfore $ deg( T_{n}) = n$ , $ [x^{n}]T_{n} = 2^{n - 1}$ , $ [x^{n - 2}]T_{n} = - 2^{n - 3}n$ ......... Let $ T_{k}(\cos \theta) = p$, $ T_{k + 1}(\cos \theta) = q$. It is easy to prove that if $ R(x) = x^{2} - 2pqx + p^{2} + q^{2} - 1$, $ R(\cos \theta) = R(\cos (2k + 1)\theta) = 0$.... It is obvious that $ R$ is a minimal degree polynomial which has root $ \cos \theta$. So $ R(x)|T_{k}(x) - p, T_{k + 1}(x) - q$, $ \cos k\theta = \cos (2k^{2} + k)\theta = p$, $ \cos (k + 1)\theta = \cos (2k + 1)(k + 1)\theta = q$, Let's Suppose that $ \frac {k(k + 1)\theta}{\pi} \not\in \mathbb{Z}$. then $ \frac {k^{2}\theta}{\pi}, \frac {(k + 1)^{2}\theta}{\pi} \in \mathbb{Z}$ $ \Rightarrow \frac {(2k + 1)\theta}{\pi}\in\mathbb{Z} \rightarrow \cos (2k + 1)\theta = \pm 1$, but $ \cos \theta$ is irrational, contradiction! Now $ \frac {k(k + 1)\theta}{\pi} \in \mathbb{Z}$. $ \Rightarrow T_{k}(q) = T_{k + 1}(p) = \pm 1$. $ q = \frac {a}{b}$ ($ (a,b) = 1$). if $ k\ge3$, then we can easily show that $ q = 0$ or $ a = \pm 1$ and $ b = \pm 2^{i}$. Therfore $ p, q = \pm \frac {1}{2^{i}}$ or $ 0$. If one can prove that $ T_{n}(\frac {1}{2^{i}})\not = \pm1$ if $ i\ge 2$, this is the end.. I think it isn't very hard.... but I don't have times to check it..

Little Gauss wrote: I don't know why I couldn't think about this...... Let $ T_{n}$ be a integer coefficient polynomial such that $ T_{n}(\cos \theta) = \cos n\theta$. It is easy to prove that $ T_{n + 1}(x) + T_{n - 1}(x) = 2xT_{n}(x)$, Therfore $ deg( T_{n}) = n$ , $ [x^{n}]T_{n} = 2^{n - 1}$ , $ [x^{n - 2}]T_{n} = - 2^{n - 3}n$ ......... Let $ T_{k}(\cos \theta) = p$, $ T_{k + 1}(\cos \theta) = q$. It is easy to prove that if $ R(x) = x^{2} - 2pqx + p^{2} + q^{2} - 1$, $ R(\cos \theta) = R(\cos (2k + 1)\theta) = 0$.... It is obvious that $ R$ is a minimal degree polynomial which has root $ \cos \theta$. So $ R(x)|T_{k}(x) - p, T_{k + 1}(x) - q$, $ \cos k\theta = \cos (2k^{2} + k)\theta = p$, $ \cos (k + 1)\theta = \cos (2k + 1)(k + 1)\theta = q$, Let's Suppose that $ \frac {k(k + 1)\theta}{\pi} \not\in \mathbb{Z}$. then $ \frac {k^{2}\theta}{\pi}, \frac {(k + 1)^{2}\theta}{\pi} \in \mathbb{Z}$ $ \Rightarrow \frac {(2k + 1)\theta}{\pi}\in\mathbb{Z} \rightarrow \cos (2k + 1)\theta = \pm 1$, but $ \cos \theta$ is irrational, contradiction! Now $ \frac {k(k + 1)\theta}{\pi} \in \mathbb{Z}$. $ \Rightarrow T_{k}(q) = T_{k + 1}(p) = \pm 1$. $ q = \frac {a}{b}$ ($ (a,b) = 1$). if $ k\ge3$, then we can easily show that $ q = 0$ or $ a = \pm 1$ and $ b = \pm 2^{i}$. Therfore $ p, q = \pm \frac {1}{2^{i}}$ or $ 0$. If one can prove that $ T_{n}(\frac {1}{2^{i}})\not = \pm1$ if $ i\ge 2$, this is the end.. I think it isn't very hard.... but I don't have times to check it.. A wondeful approach We needn't to prove $ T_{n}(\frac {1}{2^{i}})\not = \pm1$ if $ i\ge 2$ We just take your conclusion $ \frac {k(k + 1)\theta}{\pi} \in \mathbb{Z}$ We get:$ \theta$ is a rational multiple of $ \pi$ From $ R$ is a minimal degree polynomial which has root $ \cos \theta$ yields $ \cos \theta$ is a number whose algebraic degree is 2 From an old assertion of Hsiao (easy to prove) $ \theta = \frac {p}{q}\pi$ that $ q = 4,5,6$ Easy to check $ q\neq 4,5$ yields $ q = 6,p = 1$ that finished our proof

The problem statement is still true if $k+1$ is replaced by an integer $l$ relatively prime to $k$. Lemma 1 The only rational $p\in [0,1]$ such that $\cos p\pi$ is rational are $p=0, \frac{1}{3},\frac{1}{2},\frac{2}{3},1$. Proof Let $\omega=\cos (p\pi)+i\sin(p\pi)$. Then $\omega$ is a root of unity and hence an algebraic integer; similarly $\overline{\omega}$ is as well. Hence $2\cos(p\pi) =\omega + \overline{\omega}$ is an algebraic integer; since it is rational it is a rational (i.e. regular) integer. Hence we must have $2\cos(p\pi)=0,\pm 1,\pm 2$, giving $p=0, \frac{1}{3},\frac{1}{2},\frac{2}{3},1$. Lemma 2 If $\cos k\theta$ and $\cos l\theta $ are rational for relatively prime positive integers $k,l$, then either $\cos \theta $ is rational or $\theta$ is a rational multiple of $\pi$. Proof Let $\cos k\theta=p$ and $\cos l\theta =q$. Noting that $e^{i\phi}+\frac{1}{e^{i\phi}}=2\cos \phi$, we have that $e^{i\theta}$ is a root of \begin{align} x^k+\frac{1}{x^k}=2p&\implies x^{2k}-2px^k+1=0\\ x^l+\frac{1}{x^l}=2q&\implies x^{2l}-2qx^l+1=0 \end{align} Suppose $\theta$ is not a rational multiple of $\pi$. Then the numbers $e^{\pm i\theta+\frac{2\pi i j}{k}}$ for $0\leq j<k$ are all distinct. They are the $2k$ roots of (1). Similarly, $e^{\pm i\theta+\frac{2\pi i j}{l}}$ are the $2l$ roots of (2). Since $\theta$ is not a rational multiple of $\pi$ and $k,l$ are relatively prime, we have that the only solution to $e^{\pm i\theta+\frac{2\pi i j_1}{k}}= e^{\pm i\theta+\frac{2\pi i j_2}{l}}$ for $0\leq j_1<k$ and $0\leq j_2<l$ for some choices of signs is when $j_1=j_2=0$. Thus the only roots the two polynomials share in common are $\omega=e^{i\theta}$ and $\bar{\omega}=-e^{i\theta}$, and the polynomial $g(x):=(x-e^{i\theta})(x-e^{-i\theta})$ is the greatest common divisor of polynomials $x^{2k}-2px^k+1$ and $x^{2l}-2qx^l+1$, which have rational coefficients. Hence $g(x)$ has rational coefficients, and $e^{i\theta}+e^{-i\theta}=2\cos \theta$ is rational. By Lemma 2, $\theta$ is a rational multiple of $\pi$. Since by Lemma 1 the only rational multiples of $\pi$ that give a rational values for cosine are multiples of $\frac{\pi}{6}$, $k\theta$ and $l\theta$ are both multiples of $\frac{\pi}{6}$. By Bezout's Theorem $\theta=m(k\theta)+n(l\theta)$ for some integers $m,n$, so $\theta$ is a multiple of $\frac{\pi}{6}$. Since $\cos\theta$ is irrational, $\theta=\frac{\pi}{6}$. Note: This solution is based on the idea of the solution to the problem here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=589&t=290391

Let $x = \cos \theta + i \sin \theta$, and let $z_m = x^m + \tfrac{1}{x^m} = 2 \cos m\theta$ for all integer $m$. We are told that $z_k$ and $z_{k + 1}$ are both rational but $z_1$ is not. Let $T_n(z)$ be the monic polynomial with integer coefficients satisfying $T_n(2 \cos x) = 2 \cos nx$ for all $x$. Let $q(z) = \gcd(T_n(z) - z_n, T_{n + 1}(z) - z_{n + 1})$. Then $q$ has rational coefficients and $z_1$ is a root of $q$. Thus, if $\deg q = 1$, then $z_1$ is rational, contradiction. Thus, $\deg q \ge 2$. If $q$ has root $z_1$ with multiplicity $\deg q$, then it's easy to see $z_1$ is rational, contradiction. Thus, $q$ has a root $w \neq z_1$. Let $y$ be a number with $y + \tfrac{1}{y} = w$. Then since $x + \tfrac{1}{x} = z_1 \neq w = y + \tfrac{1}{y}$, $x \neq y$ and $xy \neq 1$. Now, $x^k + \tfrac{1}{x^k} = y^k + \tfrac{1}{y^k} = z_k$ and $x^{k + 1} + \tfrac{1}{x^{k + 1}} = y^{k + 1} + \tfrac{1}{y^{k + 1}} = z_{k + 1}$. Thus, we obtain $x^k = y^k$ or $(xy)^k = 1$, and $x^{k + 1} = y^{k + 1}$ or $(xy)^{k + 1} = 1$. By setting $y \to \tfrac{1}{y}$ if necessary, assume $x^k = y^k$. Then $x^{k + 1} = y^{k + 1}$ implies $x = y$, bad. Thus $(xy)^{k + 1} = 1$. Since $(x/y)^k = (xy)^{k + 1} = 1$, both $x/y$ and $xy$ are roots of unity, so $x$ and $y$ are too. It follows that angles $\alpha_1 = k \theta$ and $\alpha_2 = (k + 1) \theta$ are a rational number of degrees and have rational cosines, so \[\cos k\theta, \cos (k + 1)\theta \in \{-1, -\tfrac{1}{2}, 0, \tfrac{1}{2}, 1\}.\]It follows that $k \theta$ and $(k + 1) \theta$ are both multiples of $30^{\circ}$, so $\theta$ is too. Since $\theta \in (0^{\circ}, 90^{\circ})$, we obtain $\theta = 30^{\circ}$ or $\theta = 60^{\circ}$. Since $\cos 60^{\circ}$ is rational, it follows that $\theta = 30^{\circ}$. The solution also works for the following problem: Quote: Let $ \theta$ be an angle in the interval $ (0,\pi/2)$. Given that $ \cos \theta$ is irrational, and that $ \cos p \theta$ and $ \cos q \theta$ are both rational for some relatively prime positive integers $ p, q$, show that $ \theta = \pi/6$.

Nice problem The difficulty of the problem lies in proving that $\theta$ is a rational multiple of $\pi$. Suppose otherwise, and let $z=\operatorname{cis}(\theta)$. Then since $\cos k\theta:=q_1$ is rational, we have $$\operatorname{Re}(z^k)=q_1 \implies z^k+z^{-k}=2q_1 \implies z^{2k}-2q_1z^k+1=0,$$so $z$ is a root of $P_1(x):=x^{2k}-2q_1x^k+1$. Likewise, $z$ is a root of $P_2(x):=x^{2k+2}-2q_2x^{k+1}+1$, where $\cos (k+1)\theta=q_2$, so it's a root of $\gcd(P_1(x),P_2(x)) \in \mathbb{Q}[x]$. It turns out that we can find all roots of $P_1(x)$ without great difficulty: observe that all numbers of the form $\operatorname{cis}(\pm\theta+\tfrac{2n\pi}{k})$ are roots, where $n=0,\ldots,2k-1$. Since there are $2n$ of these (all such roots are distinct because $\theta$ isn't a rational multiple of $\pi$), they are actually precisely the roots of $P_1$. Likewise, the roots of $P_2$ are precisely $\operatorname{cis}(\pm \theta+\frac{2n\pi}{k+1})$. Because $\theta$ isn't a rational multiple of $\pi$ and $k \perp k+1$, the only shared roots between $P_1$ and $P_2$ are $\operatorname{cis}(\pm\theta)=z,\overline{z}$. Thus $$\gcd(P_1(x),P_2(x))=(x-z)(x-\overline{z}) \in \mathbb{Q}[x],$$which implies that $z+\overline{z} \in \mathbb{Q} \implies \operatorname{Re}(z)=\cos \theta \in \mathbb{Q}$: illegal. Thus $\theta$ must be a rational multiple of $\pi$. By Niven's Theorem, if $\phi$ is a rational multiple of $\pi$ and $\cos(\phi) \in \mathbb{Q}$, then $\phi \in \{0,\pi/3,\pi/2,2\pi/3\} \pmod{\pi}$, so modulo $\pi$ the only possible differences between two rational multiples of $\pi$ whose cosines are rational are $0,\pi/6,\pi/3,\pi/2,2\pi/3$. Since the cosines of $k\theta$ and $(k+1)\theta$ are both rational, it follows that $$(k+1)\theta-k\theta=\theta \in \{0,\pi/6,\pi/3,\pi/2,2\pi/3\} \pmod{\pi},$$so if $\theta \in (0,\pi/2)$ then we must have $\theta \in \{\pi/6,\pi/3\}$. But $\cos \pi/3$ is rational, hence $\theta=\pi/6$ as desired. $\blacksquare$

Here we go: If $\theta$ is a rational multiple of $\pi$, by Niven's theorem, $k \theta , (k+1) \theta \in \{0,\pi/3,\pi/2,2\pi/3\} \pmod{\pi} $ Thus, their difference $(k+1) \theta - k \theta = \theta \in \{0,\pi/6,\pi/3,\pi/2,2\pi/3\} \pmod \pi$. Since $\theta \in (0, \pi/2)$, and $\cos \theta$ is irrational, $\theta = \pi/6$. If $\theta$ is not a rational multiple of $\pi $: Consider the following two polynomials: $$P(x)= x^{2k}-2 \cos (k \theta) x^k+1, G(x) = x^{2(k+1)}-2 \cos ((k+1) \theta) x^{k+1}+1.$$Their roots are respectively: $$z = \text{cis}\left( \pm \theta + \frac{2\pi s}{k} \right)$$$$ z = \text{cis} \left( \pm \theta + \frac{2\pi t}{k+1} \right).$$Since $\theta$ is not a rational multiple of $\pi$: the only common roots are $\text{cis} ( \pm \theta )$. Thus: $H(x) = \gcd(P, G) = (x-\alpha)(x-\bar{\alpha})$ where $\alpha = \text{cis}(\theta)$. Since $P, G \in \mathbb Q [x ]$, we have: $H \in \mathbb Q [ x ]$. Thus: $\alpha + \bar{\alpha} = 2 \cos \theta \in \mathbb Q$ which contradicts that $\cos \theta$ is irrational.