Is $\frac{\sqrt3+\sqrt5}{2}$ an algebraic integer?

Question

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?

Answer 1

To go from $\alpha^4$ to $\left( \dfrac{\alpha}{2} \right)^4$, you have to divide by $2^4=16$; the equation you highlight is the result of dividing through the previous equation by $16$, writing it as a polynomial in $\dfrac{\alpha}{2}$, and making the powers of $2$ involved in the division explicit.

Answer 2

Take $\alpha^4-16\alpha^2+4=0$ and write $2\cdot\frac\alpha2$ instead of $\alpha$ everywhere. Now divide everything by the leading coefficient $2^4$.

Answer 3

\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}