<s>[INST] Is $\frac{\sqrt3+\sqrt5}{2}$ an algebraic integer? 
Find the minimal polynomial of $\sqrt3+\sqrt5$ over $\mathbb Q$. Is $\Large \frac{\sqrt3+\sqrt5}{2}$ an algebraic integer?

The first part is fine and I found that the minimal polynomial of $\sqrt3+\sqrt5$ over $\mathbb Q$ is $x^4-16x^2+4$. I have some questions about the written answer.
Let $\alpha = \sqrt3+\sqrt5 $ so will try to find if $\frac{\alpha}{2}$ is algebraic integer.
The real answer:
$\alpha^4-16\alpha^2+4=0$

$\Large (\frac{\alpha}{2})^4-(\frac{16}{2^2})(\frac{\alpha}{2})^2+(\frac{4}{2^4})=0$

$\Large (\frac{\alpha}{2})^4-4(\frac{\alpha}{2})^2+(\frac{1}{4})=0$
Minimal polynomial of $\Large \frac{\alpha}{2}$
is $x^4-4x^2+\frac{1}{4}$, which is not defined over $\mathbb Z$ hence $\Large \frac{\alpha}{2}$ is not an algebraic integer.
I've highlighted the line which confuses me. Can anyone show me where this comes from?
I thought they may have substituated $\alpha$ with $\frac{\alpha}{2}$ in the minimal polynomial but then I don't understand where $\frac{16}{2^2}$ and $\frac{4}{2^4}$ came from?
 [/INST] \begin{align}\alpha^4-16\alpha^2+4=0&\iff\frac{\alpha^4-16\alpha^2+4}{2^4}=0\\&\iff\left(\frac\alpha2\right)^2-4\left(\frac\alpha2\right)^2+\frac14=0.\end{align}
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