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?