algebra precalculus - Polynomial p(a)=1, why does it have at most 2 integer roots?

Suppose is p(x) is a polynomial with integer coefficients. Show that if p(a)=1 for some integer a then p(x) has at most two integer roots.

I have read the answers given in an earlier post but don't understand why the proof in the answer is complete.

answer on the other post

If x1,x2,x3 are three distinct integer roots of p, we can write
p(x)=(x−x1)(x−x2)(x−x3)q(x)
and find
1=(a−x1)(a−x2)(a−x3)q(a)
where a−x1,a−x2,a−x3 are pairwise distinct integers and q(a) is a nonzero integer. At most two of a−x1,a−x2,a−x3 can be ∈{±1}.

My question is, how do you know that q(a) is an integer? Why couldn't q(a) be equal to, say 1/2, and one of the other factors equal to 2, therefore giving other integer solutions?

Answer

This is a consequence of various important facts about the integers and integer polynomials. Suppose $p(x)$ has integer coefficients and $p(x_1)=0$ for some integer $x_1$. We can use polynomial division to obtain $$p(x)=(x-x_1)q_1(x)+r_1$$

Now because $x-x_1$ is monic (the coefficient of $x$ is equal to $1$), the division will never result in a non-integer fraction, so that $q_1(x)$ will have integer coefficients, and $r_1$ will be an integer. Setting $x=x_1$ we find that $r_1=0$.

Now $p(x_2)=(x_2-x_1)q_1(x_2)=0$ and since $x_2\neq x_1$ and the integers have no non-trivial zero divisors - they are an integral domain - we must have $q_1(x_2)=0$. Hence $q_1(x)=(x-x_2)q_2(x)$ by the same argument as with $p(x)$ and $q_2(x)$ has integer coefficients.

So when you conclude the argument for the form of $p(x)$ you can conclude that $q(x)$ has integer coefficients.

In later work you will find the notions of "integer" and "monic polynomial" inextricably linked (an algebraic integer satisfies a monic polynomial with integer coefficients). That linking is related to this simple fact about division, and has deeper consequences too.