Determine the minimal polynomial of $\sqrt 3+\sqrt 5$ I am struggling in finding the minimal polynomial of $\sqrt{3}+\sqrt{5}\in \mathbb C$ over $\mathbb Q$
Any ideas? I tried to consider its square but it did not helped..

As was mentioned in the other answers, $x^4-16x^2+4$ is a polynomial with root $\sqrt{3}+\sqrt{5}$, so it suffices to prove that this polynomial is irreducible. There are several methods:
1) Calculation of the quadratic factors, see Lubin's answer.
2) Field theory: $\mathbb{Q}(\sqrt{3},\sqrt{5}) = \mathbb{Q}(\sqrt{3})(\sqrt{5})$ has degree $2 \cdot 2=4$ over $\mathbb{Q}$ essentially because $\sqrt{5} \notin \mathbb{Q}(\sqrt{3})$ (easy computation!) and hence $X^2-5$ stays irreducible over $\mathbb{Q}(\sqrt{3})$. So it suffices to prove that $\mathbb{Q}(\sqrt{3},\sqrt{5})=\mathbb{Q}(\sqrt{3}+\sqrt{5})$ (because then $\sqrt{3}+\sqrt{5}$ has degree $4$ over $\mathbb{Q}$). Since $\supseteq$ is trivial, it suffices to check $\subseteq$. Notice that rationalisiation yields
$$\frac{1}{\sqrt{3}+\sqrt{5}}=\frac{\sqrt{3}-\sqrt{5}}{3-5}.$$
Hence, $\sqrt{3}-\sqrt{5} \in \mathbb{Q}(\sqrt{3}+\sqrt{5})$. But then we also have $2 \sqrt{3} = (\sqrt{3}-\sqrt{5}) + (\sqrt{3}+\sqrt{5}) \in \mathbb{Q}(\sqrt{3}+\sqrt{5})$. This shows $\sqrt{3} \in \mathbb{Q}(\sqrt{3}+\sqrt{5})$ and then we also get $\sqrt{5} \in \mathbb{Q}(\sqrt{3}+\sqrt{5})$. $\checkmark$
The same proof shows: If $p,q$ are distinct prime numbers, then $\mathbb{Q}(\sqrt{p},\sqrt{q})=\mathbb{Q}(\sqrt{p}+\sqrt{q})$ has degree $4$ over $\mathbb{Q}$. The minimal polynomial of $\sqrt{p}+\sqrt{q}$ is $(x^2-p-q)^2 - 4 pq$.
3) Kummer theory: It is a special case of Kummer theory that for distinct prime numbers $p_1,\dotsc,p_n$ (actually pairwise coprime square-free integers also work) $\mathbb{Q}(\sqrt{p_1},\dotsc,\sqrt{p_n})$ has degree $2^n$ over $\mathbb{Q}$, has primitive element $\alpha=\sqrt{p_1} + \dotsc + \sqrt{p_n}$ and that its minimal polynomial therefore has degree $2^n$. It is given by the product of all $X-\beta$, where $\beta$ runs through the conjugates $\pm \sqrt{p_1} \pm \dotsc \pm \sqrt{p_n}$. (It is quite obvious that this polynomial is rational and has $\alpha$ as a root, but its irreducibility is not so easy to prove directly, and this is where Kummer theory helps.) 
