What is the minimal polynomial of $\sqrt 3+\sqrt 5$ over $\mathbb Q(\sqrt 5)$? What is the minimal polynomial of $\sqrt 3+\sqrt 5$ over $\mathbb Q(\sqrt 5)$ ?
I proved that $\mathbb Q(\sqrt 3+\sqrt 5)=\mathbb Q(\sqrt 3,\sqrt 5)\supset \mathbb Q(\sqrt 5)$, and thus that $\mathbb Q(\sqrt 3+\sqrt5 )/\mathbb Q(\sqrt 5)$ has degree $2$. Since the minimal polynomial of $\sqrt 3+\sqrt 5$ is $X^4-16X^2+4$, the minimal polynomial of $\sqrt 3+\sqrt 5$ over $\mathbb Q(\sqrt 5)$ must be a polynomial of order 2 that divide $X^4-16X^2+4$. It's roots are $\pm\sqrt{8\pm 2\sqrt{15}}$. But I tried to make all combinaisons $(X\pm\sqrt{8\pm2\sqrt{15}})(X\mp\sqrt{8\mp2\sqrt{15}})$, but I didn't get any polynomial of order $2$ in $\mathbb Q(\sqrt 5)$ that canceled $\sqrt 3+\sqrt 5$. Where is my mistakes ? 

Its conjugate is $\sqrt5-\sqrt3$.
Therefore, the minimal polynomial is:
$$\begin{array}{rcl}
(x-(\sqrt3+\sqrt5))(x-(\sqrt5-\sqrt3))
&=& (x-\sqrt5)^2 - (\sqrt3)^2 \\
&=& x^2-2\sqrt5x+5-3 \\
&=& x^2-2\sqrt5x+2
\end{array}$$
