About algebraic integer and algebraic number 
*

*Let $u$ a algebraic number. Prove that exists a natural number $n\in \mathbb{Z}$ such that $nu$ is a algebraic integer 

*If $u$ is algebraic integer  and $n\in \mathbb{Z}$ then $u+n$ and $nu$ are algebraic integers. 


I don't see how can I start. 
Remember that: $u$ is algebraic integer if it is a root by a monic polynomial $f(x)\in \mathbb{Z}[x]$.  
 A: An algebraic number satisfies an equation like
$$
P(x)=a_kx^k+a_{k-1}x^{k-1}+a_{k-2}x^{k-2}+\dots+a_1x+a_0=0\tag{1}
$$
where $a_j\in\mathbb{Z}$ and $a_k\not=0$.
An algebraic integer satisfies $(1)$ with $a_k=1$.
Hint for 1.
Suppose $u$ satisfies $(1)$ but $a_k\not=1$. Figure out what polynomial $a_ku$ satisfies (and cancel all common factors that you can easily find).
Hint for 2.
Suppose $u$ satisfies $(1)$ with $a_n=1$. Use the binomial theorem to figure out what polynomial $u+n$ satsfies (further hint: note that $P((u+n)-n)=0$). Also, figure out what polynomial $nu$ satisfies (another further hint: note that $n^kP((nu)/n)=0$ and cancel all common factors that you can easily find).
A: $(i)$ Let $P(X)=a_mX^m+\cdots+a_1X+a_0 \in \mathbb{Z}[x]$ be such that $P(u)=0$.
Then $a_mu$ is a root of the polynomial.....
$(ii)$. If $P(X)=X^m+a_{m-1}X^{m-1}+\cdots+a_0$ is a polynomial so that $P(u)=0$, then $u+n$ is a root of $P(X-n)$. Also $n^mP(u)=Q(nu)$ for some polynomial $Q$.
A: Let me give you a hint for the first part. Let's take a concrete example here. Suppose that $x$ satisfies the monic polynomial equation 
$$
x^2 - \frac{3}{4}x + \frac{1}{5} = 0
$$
Then you want to prove that there exists an integer $n$ such that $nx$ satisfies a similar monic polynomial equation but with integer coefficients. Well the idea is that you can clear denominators. For example since $4$ and $5$ appear as denominators we can multiply by $20$ to get
$$
20x^2 - 15x + 4 = 0
$$
But now the problem is that this is no longer monic. But then we can multiply again by 20 so that we get a factor $20^2$ in the leading coefficient which then can be absorbed by the square as follows
$$
20^2 x^2 - 15\cdot 20x + 4\cdot20 = 0 \implies (\color{red}{20x})^2 -15 \cdot(\color{red}{20x}) + 4\times 20 = 0
$$
so you see that now $\alpha = 20x$ satisfies
$$
\alpha^2 - 15\alpha + 80 = 0
$$
Thus $\alpha = 20x$ is an algebraic integer. And this same idea works in general.
