if $a$ is an algebraic integer and $m\in \mathbb{Z}$ then $a+m$ is an algebraic integer.

Question

I tried to use newton binomial, if $p_a(x) = \sum_0^nb_kx^k$ , then $b_k(a+m)^k =b_ka^k +b_k\sum_0^{k-1}a^im^{k-i}$, denote $T_k=b_k\sum_0^{k-1}a^im^{k-i}$, one gets that $p_{a}(x)-(\sum_1^nT_k ) - mb_0$ has $(m+a)$ as a root, but the problem is that $T_k$ are not guaranteed to be integers. I also worked out the $a\cdot m$ must be an algebraic integer (by taking $p_a(x)=\sum_0^nb_kx^k$ and define: $\sum_0^n m^{n-k}b_kx^k$ ), yet I didn't succeed to use it in order to manipulate $p_{a}(x)-(\sum_1^nT_k ) - mb_0$ back to $\mathbb{Z}[x]$

Answer 1

Hint: If $P(x)=\sum_{k=0}^n b_k X^k$ has $a$ as a root, then

$$P(X-m)=\sum_{k=0}^n b_k (X-m)^k$$ has $a+m$ as a root. As a polynomial this does have integer coefficients.

Your mistake was thinking about this as an expression in $a$ not as a polynomial. You need the coefficient of $a^j$ to be an integer, not $T_k$.