Let me complete the solution: Firstly,if $\tan x+\tan y=2$,then \[\sin(x+y)=2\cos x\cos y=cos(x+y)+\cos(x-y)\Longrightarrow \sqrt{2}(\sin(x+y-\frac{\pi}{4}))=\cos(x-y) \Longrightarrow 2(\sin(x+y-\frac{\pi}{4}))^{2}=1-\sin^{2}(x-y) \Longrightarrow \cos(2(x+y)-\frac{\pi}{2})=\cos^{2}(x-y)\] Let $z=x-y,t=2(x+y)-\frac{\pi}{2}$ then $\cos^{2}z=\cos t$ Then we only need to find all pairs of (z,t) such that $z,t$ both are rational multiples of $\pi$ and $\cos^{2}z=\cos t$ Lemma1,Define ${p_{n}(x)=\prod_{1\le k \le \frac{n}{2},gcd(k,n)=1}(x-\cos \frac{2k\pi}{n})},(n\ge 3)$ then we firstly proved $p_{n}(x)$ is an polynomial with rational coeffients,and it is irreducible(Do I spell correctly?I have not a dictionary here...). Proof of Lemma 1:Let $f_{n}(x)$ be the polynomial with the property $f_{n}(\cos x)=\cos nx$. then we can prove that $f_{2n+1}(x)-1=(x-1)h_{2n+1}(x)^{2},f_{2n}=(x-1)(x+1)h_{2n}(x)^{2}$ where $h_{n}(x)=\prod_{1\le k< \frac{n}{2}}{(x-\cos{\frac{2k\pi}{n})}}$ and we can easily find that $\prod_{d|n}{p_{d}(x)}=h_{n}(x)$ then $p_{n}(x)=\prod_{d|n}{(h_{d}(x))^{\mu(\frac{n}{d})}}$ is a polynomial. We consider ${p_{n}(\frac{t+\frac{1}{t}}{2})=\prod_{1\le k \le \frac{n}{2},gcd(k,n)=1}(\frac{t+\frac{1}{t}}{2}-\cos \frac{2k\pi}{n})}=$,$\frac{\Phi_{n}(t)}{t^{\frac{\phi(n)}{2}}}$ becuase $\Phi_{n}(t)$ is irreducible,$p_{n}(x)$ is also irreducible. Lemma 2IF a polynomial $f(x)\in Q[x]$ is irreducible, and $f(x^{2})$ is reducible,then there exist $g(x)\in Q[x]$ such that $f(x^{2})=\pm g(x)g(-x)$. Suppose that $z=\frac{u}{v}\pi ,t=\frac{m}{n}\pi,(n,v\ge 3)$ and Let $q(x)=p_{n}(x^{2})$,then $p_{v}(\cos z)=0,p_{n}(\cos t)=0,q(\cos z)=0$ by Lemma 1 ,$p_{v}(x)|q(x)$ by Lemma 2,$p_{v}(x)=q(x) ,or, p_{v}(x)p_{v}(-x)=\pm q(x)$ But both of these two case leads to contraduction,becuase $q(x)$ has at least one non-real root,but $p_{v}(x)$ does not. I hope my solution is correct.Could somebody tell me if it is correct?I am not so confident....... @Peter,could you help me delet the first two post?

Hawk Tiger wrote: @Peter,could you help me delet the first two post? I have deleted the posts :smile: ,though I am not Peter .

Hawk Tiger wrote: Let me complete the solution: Firstly,if $ \tan x + \tan y = 2$,then \[ \sin(x + y) = 2\cos x\cos y = cos(x + y) + \cos(x - y)\Longrightarrow \sqrt {2}(\sin(x + y - \frac {\pi}{4})) = \cos(x - y) \Longrightarrow 2(\sin(x + y - \frac {\pi}{4}))^{2} = 1 - \sin^{2}(x - y) \Longrightarrow \cos(2(x + y) - \frac {\pi}{2}) = \cos^{2}(x - y) \] Let $ z = x - y,t = 2(x + y) - \frac {\pi}{2}$ then $ \cos^{2}z = \cos t$ Then we only need to find all pairs of (z,t) such that $ z,t$ both are rational multiples of $ \pi$ and $ \cos^{2}z = \cos t$ Lemma1,Define ${ p_{n}(x) = \prod_{1\le k \le \frac {n}{2},gcd(k,n) = 1}(x - \cos \frac {2k\pi}{n})},(n\ge 3)$ then we firstly proved $ p_{n}(x)$ is an polynomial with rational coeffients,and it is irreducible(Do I spell correctly?I have not a dictionary here...). Proof of Lemma 1:Let $ f_{n}(x)$ be the polynomial with the property $ f_{n}(\cos x) = \cos nx$. then we can prove that $ f_{2n + 1}(x) - 1 = (x - 1)h_{2n + 1}(x)^{2},f_{2n} = (x - 1)(x + 1)h_{2n}(x)^{2}$ where $ h_{n}(x) = \prod_{1\le k < \frac {n}{2}}{(x - \cos{\frac {2k\pi}{n})}}$ and we can easily find that $ \prod_{d|n}{p_{d}(x)} = h_{n}(x)$ then $ p_{n}(x) = \prod_{d|n}{(h_{d}(x))^{\mu(\frac {n}{d})}}$ is a polynomial. We consider ${ p_{n}(\frac {t + \frac {1}{t}}{2}) = \prod_{1\le k \le \frac {n}{2},gcd(k,n) = 1}(\frac {t + \frac {1}{t}}{2} - \cos \frac {2k\pi}{n})} =$,$ \frac {\Phi_{n}(t)}{t^{\frac {\phi(n)}{2}}}$ becuase $ \Phi_{n}(t)$ is irreducible,$ p_{n}(x)$ is also irreducible. Lemma 2IF a polynomial $ f(x)\in Q[x]$ is irreducible, and $ f(x^{2})$ is reducible,then there exist $ g(x)\in Q[x]$ such that $ f(x^{2}) = \pm g(x)g( - x)$. Suppose that $ z = \frac {u}{v}\pi ,t = \frac {m}{n}\pi,(n,v\ge 3)$ and Let $ q(x) = p_{n}(x^{2})$,then $ p_{v}(\cos z) = 0,p_{n}(\cos t) = 0,q(\cos z) = 0$ by Lemma 1 ,$ p_{v}(x)|q(x)$ by Lemma 2,$ p_{v}(x) = q(x) ,or, p_{v}(x)p_{v}( - x) = \pm q(x)$ But both of these two case leads to contraduction,becuase $ q(x)$ has at least one non-real root,but $ p_{v}(x)$ does not. I hope my solution is correct.Could somebody tell me if it is correct?I am not so confident....... @Peter,could you help me delet the first two post? Thank you for brilliant lemma 1. BUT I don't think $ Q(X) = \pm P_v(X)P_v( - X)$ Lemma ONLY tell you $ P_v(X)P_v( - X)$ divides $ Q(X)$ Maybe $ Q(X) = P_v(X)P_v( - X)R(X)R( - X)$ Right?

hxy09 wrote: Lemma ONLY tell you $ P_v(X)P_v( - X)$ divides $ Q(X)$ Maybe $ Q(X) = P_v(X)P_v( - X)R(X)R( - X)$ Thenyou write $ Q(X) = \big(P_v(X)R(X)\big)\big(P_v( - X)R( - X)\big)$, so his claim for $ g(X)=P_v(X)R(X)$? (Your claim misses the $ \pm$ though)

Peter wrote: hxy09 wrote: Lemma ONLY tell you $ P_v(X)P_v( - X)$ divides $ Q(X)$ Maybe $ Q(X) = P_v(X)P_v( - X)R(X)R( - X)$ Thenyou write $ Q(X) = \big(P_v(X)R(X)\big)\big(P_v( - X)R( - X)\big)$, so his claim for $ g(X) = P_v(X)R(X)$? (Your claim misses the $ \pm$ though) Sorry I think you didn't really catch my question.Lemma 2 is obviously true.What I mean is $ q(x)$ has at least one non real root but $ p_v(x)$ doesn't.HOWEVER it doesn't lead to contradiction.You see, $ R(x)$ and$ R( - x)$ might have non real root.

I'm not sure I understand you. His argument (at least how I interpret it) is that $ p_v$ is irreducible, $ p_v|q$, and $ \deg(q)=2\deg p_v$ by definition. Hence if $ g(x)=p_v(x)R(x)$, then $ \deg(R)=0$...

Peter wrote: I'm not sure I understand you. His argument (at least how I interpret it) is that $ p_v$ is irreducible, $ p_v|q$, and $ \deg(q) = 2\deg p_v$ by definition. Hence if $ g(x) = p_v(x)R(x)$, then $ \deg(R) = 0$... Sorry,but I think $ \deg(q) = 2\deg p_n$ by definition not $ \deg(q) = 2\deg p_v$ right?

Solution:Take derivtive to find $tanx+tany$ is monotonic increasing as $x$ or $y$ increases. So the only solution is $x=y=\frac {\pi}{4}$