Explanation of solution of a question about polynomials Taken from https://artofproblemsolving.com/wiki/index.php/1970_Canadian_MO_Problems/Problem_10
Problem 
Given the polynomial $f(x)=x^n+a_{1}x^{n-1}+a_{2}x^{n-2}+\cdots+a_{n-1}x+a_n$ with integer coefficients $a_1,a_2,\ldots,a_n$, and given also that there exist four distinct integers $a, b, c$ and $d$ such that $f(a)=f(b)=f(c)=f(d)=5$, show that there is no integer $k$ such that $f(k)=8$.  
Solution 
Set $g(x) = f(x) − 5$. Since $a, b, c, d$ are all roots of $g(x)$, we must have
$g(x) = (x − a) (x − b) (x − c) (x − d) h(x)$
for some $h(x) ∈ Z[x].$
... 
How do we know that h(x) has integral coefficients here? How is it obvious? 
I tried simplifying the $g(x)$ expression and comparing coefficients, and it seems to be true, but cannot prove it. Moreover, how is it so obvious?
 A: *

*If $n<4$ then $g=0$ because $g(x)=0$ when $x$ is any of the 4 distinct values $a,b,c,d.$


*If $n\ge 4$ let $j(x)=(x-a)(x-b)(x-c)(x-d).$
There exist $h(x), s(x)\in \Bbb Z[x]$ with $g(x)=h(x)j(x)+s(x)$ and $deg(s)<n$ because $deg (\,g(x)-x^{n-4}j(x)\,)<n.$
So consider $m,$ the least degree of an $s(x)\in \Bbb Z[x]$ such that there exists  $h(x)\in \Bbb Z[x]$ with $g(x)=h(x)j(x)+s(x).$
Suppose $m\ge 4.$ Let  $g(x)=h_1(x)j(x)+s_1(x)$ with $h_1(x)$ and $s_1(x)$ in $\Bbb Z[x]$ and $deg(s_1)=m.$ Let the leading term of $s_1(x)$ be $\sigma x^m.$  Let $s_2(x) =s_1(x)-\sigma x^{m-4}j(x).$ Let $h_2(x)=h_1(x)+\sigma x^{m-4}.$ Then $h_2(x)$ and $s_2(x)$ are in $\Bbb Z[x]$ with $g(x)=h_2(x)j(x)+s_2(x)$ and $deg (s_2)\le m-1,$ contradicting the minimality of $m.$
So by contradiction we have $m<4.$
So let $g(x)=h(x)j(x)+s(x)$ with $h(x)$ and $s(x)$ in $\Bbb Z[x]$ and $deg (s)<4.$ But $s(x)=0$ when $x$ is any of the 4 different values $a,b,c,d$ and $deg (s)<4$. So $s=0$ and $g(x)=h(x)j(x).$
A: Let $f(x)=\sum_{k=0}^n a_kx^k $ with $a_k\in\mathbb Z $ have a root $A $. Then by FTA:
$$
g (x)=\frac {f (x)}{x-A}=\sum_{k=0}^{n-1} b_kx^k
$$
is a polynomial of degree $n-1 $. By polynomial division one obtains:
$$b_{n-1}=a_n;\quad b_i=a_i+A\times b_{i+1}\ (n-2\ge i\ge0).$$
Since integer numbers are closed under operations of addition and multiplication one concludes: $$A  \in\mathbb Z\implies  b_k\in\mathbb Z .$$
The same conclusion is valid by induction in the case of several integer roots.
