Factoring polynomials in $\Bbb Z[x]$ by interpolating at factors of their values

Question

Background

Exercise 17: This exercise shows that factorization in $\Bbb{Q}[x]$ is algorithmic. Let $f(x)\in \mathbb{Z}[x]$ be a polynomial degree $n$. If $f$ is reducible, it has a factor $g$ of degree at most $\lfloor\frac{n}{2} \rfloor$. Take any integers $a_1,\ldots,a_{{\lfloor\frac{n}{2} \rfloor}+1}$. We have that $g(a_i)$ divides $f(a_i)$, so there are only a finite number of possibilities for $g(a_i)$. A polynomial of degree $k$ is determined by its values in $k+1$ points. Thus there are only finitely many possible factors $g$ to test.

Questions

For the above exercise, there are several things that are confusing. The statements:

If $f$ is reducible it has a factor $g$ of degree at most $\lfloor\frac{n}{2} \rfloor\quad (1)$.

Take any integers $a_1,\ldots,a_{{\lfloor\frac{n}{2} \rfloor}+1}\quad (2)$.

A polynomial of degree $k$ is determined by its values in $k+1$ points.

For the statement stated in $(1)$ is that a theorem, I am not sure where did $\lfloor\frac{n}{2} \rfloor$ come from.

For $(2)$, assuming $(1)$ to be true, ${\lfloor\frac{n}{2} \rfloor}+1$ why do we need ${\lfloor\frac{n}{2} \rfloor}+1$ $a_i$? Could it just be as well if we need $i+1$ for $i\in \Bbb{N}$?

For $(3)$ I thought a polynomial of degree $k$ has at most $k$ roots, so we only need $k$ number of points? Or is the author referring to some other math facts?

Answer 1

(1): Yes, that's a theorem. If we have two integer polynomials of degree $i$ and $j$, their product has degree $i+j$. If $f$ is reducible, then there are two polynomials whose product is $f$, and $f$ has degree $n$. So if $i$ and $j$ are the degrees of these two factor polynomials, we have $i+j = n$. And the smallest one of $i$ and $j$ can be at most $\frac n2$ (they can't both be larger than $\frac n2$), and it must also be a natural number. Which is to say, it can be at most $\left\lfloor \frac n2\right\rfloor$.

(3): Yes, a polynomial of degree $k$ has at most $k$ roots. But you need one more point to uniquely determine the polynomial. For instance, you need two points to determine a line ($k = 1$). And if you know all the roots of a polynomial, then you know a lot about the polynomial, but you still know nothing about the magnitude of the leading coefficient, so you need one more point of data.

In light of (1), if we can check every single degree-$\left\lfloor \frac n2\right\rfloor$-or-less polynomial and see whether it's a factor, we can determine whether $f$ is irreducible. Of course, there are infinitely many of them, so we can't do that directly. However, given that $g\mid f$, we must have that $g(a)\mid f(a)$ for every single integer $a$. That means we can rule out a whole bunch of candidates immediately. For instance, we must have $g(0)\mid f(0)$, so instead of infinitely many possible constant terms to check, we now have a finite number of possible constant terms. And each new $a_i$ we bring in will have a similar reducing effect on the space of candidate polynomials we actually have to check.

In light of (3), we know that if we decide on one possible combination for all the $\left\lfloor \frac n2\right\rfloor + 1$ different values $g(a_i)$, then there is exactly one degree-$\left\lfloor \frac n2\right\rfloor$-or-less polynomial that has those values at those points. And we know there are only finitely many such combinations to check. And we know that if $f$ actually has a degree-$\left\lfloor \frac n2\right\rfloor$-or-less factor, then we must hit that factor with our search.