I showed by a direct calculation that $ \alpha_1 =\sqrt 2 + \sqrt 3$ is an algebraic integer of $p(x) = x^4 -10x^2 +1$.
In the next section, I was asked to infer that $-\sqrt 2 - \sqrt 3, -\sqrt 2 + \sqrt 3, \sqrt 2 - \sqrt 3$ are also algebraic integers of $p(x)$.
I can deduce that $\alpha_2 = -(\sqrt 2 + \sqrt 3)$ is an algebraic integer too since the exponents of $p(x)$ are all even (= it's an even function)
How can I infer that the other are algebraic integers too (I don't think I'm expected to explicitly calculate it)
$\alpha=\sqrt{2}+\sqrt{3}$ is an algebraic integer over $\mathbb{Q}$ since it is a root of $p(x)=x^4-10x^2+1$.
Indeed $\alpha^2=5+2\sqrt{6}$, so $\alpha$ is a root of $(x^2-5)^2-24 = p(x)$.
But the same argument applies to $\pm\sqrt{2}\pm\sqrt{3}$, so these four numbers are all the roots of $p(x)$.
Once you prove that $p(x)$ is irreducible over $\mathbb{Q}$, for instance by showing that $p(x)$ has no linear or quadratic factor in $\mathbb{Q}[x]$, you have that $p(x)$ is the minimal polynomial of $\alpha$ over $\mathbb{Q}$ and $\pm\sqrt{2}\pm\sqrt{3}$ are algebraic conjugates.