Given a monic polynomial $f(x)$ of degree $n$ over $Z$ and $k, p \in N$ , prove that if none of the numbers $f(k), f(k + 1),..., f(k + p)$ is divisible by $p + 1$, then $f(x) = 0$ has no rational solution.
Question from 'Problem solving Strategies'.
Here is the solution
Can anyone please explain why the polynomial $g(x)$ should have integral coefficients (2nd line of solution)?
You can use division algorithm since $x-m$ has leading coefficient 1.
Alternatively, from division algorithm of $\mathbb R$ and factor theorem you can write in that way where $g(x)$ has real coefficients. Now suppose $g(x)=b_0+b_1x+...+b_nx^n$ consider $f(x)= a_0+...+a_{n+1}x^{n+1}$ where $a_i\in \mathbb Z, \forall i$.
Now compute the product.
The leading term is $b_n=a_{n+1}\in \mathbb Z$ now $a_n= b_{n-1}-mb_n \in \mathbb Z \Rightarrow b_{n-1} \in \mathbb Z$ use induction on $n$ to complete the proof.
The key point is that the division algorithm for polynomials, which in principle works fine over fields, can be done in general for any ring if the polynomial you divide for is monic.
In the case you are interested, which you divide by $x-m$, one can use Ruffini's rule to get that for any polynomial $p(x)$ with coefficients over a ring $R$ and any $m\in R$, there is a polynomial $g(x)$ with coefficients in $R$ such that $$p(x)=(x-m)q(x)+p(m).$$ The coefficients of $g(x)$ are obtained recursively: if $p(x)=a_nx^n+\cdots+a_0$ and $g(x)=b_{n-1}x^{n-1}+\cdots +b_0$, we have $b_n=a_n$ and $b_i=b_{i+1}\cdot m+a_{i+1}$.
