Given an integer-valued polynomial, can its sequential values be used to determine whether it is reducible?

Let $p(x)$ be an integer-valued polynomial of degree $n$. Is it possible to use up to $n$ consecutive values to potentially identify $p(x)$ as irreducible over the set of integer-valued polynomials?

I managed to prove that for $n=2$ there is a definite upper limit for the 3rd sequential value, but it is very difficult for higher degrees, even with induction. I encountered this question while considering some primes appearing in polynomial sequences. I have attempted to use Lagrange interpolation and binomial forms and I haven't made any progress past $n=2$.

Here is a sketch for the proof that there is an upper bound for the described polynomial with $n=2$:

We have fixed positive integers $v_1,v_2, v_1\le v_2$ and polynomial in binomial form

$$p(x) = a{x\choose 2}+(v_2-v_1)x+v_1$$

(Notice that we have set $p(0) = v_1,p(1) = v_2$.) Since we have $n=2$, we must have $p(2) = v_3$ for whatever positive integer value $v_3$ takes on, which means that we have $p(2) = a + 2v_2-v_1 = v_3$, and so we have $a = v_3-2v_2+v_1$ (which is correct by Finite Difference theory), giving us

$$p(x) = (v_3-2v_2+v_1){x\choose 2}+(v_2-v_1)x+v_1$$

This shows that only the (binomial-form) quadratic term will change in coefficient as $v_3$ changes. Now consider this value:

$$p\left(\frac 12\right)=\frac 14v_1+v_2-\frac 14v_3$$

We again have linear change in the value at this point as $v_3$ changes, and it is clear that it is possible for this value to be negative.

If $p(x)$ is composite with at least two non-constant integer-valued polynomial factors, and if $v_3$ is large enough to force $p\left(\frac 12\right)$ to be negative, then these factors are based exactly on the roots between $p(0)$ and $p(1)$, since each of these values is positive. So we would have

$$p(x) = (v_3-2v_2+v_1)(x-q_0)(x-q_1)=a(x-q_0)\cdot b(x-q_1)$$

In order for these factors to be considered "integer-valued", both $a$ and $b$ must be large enough that either the denominator of the rational $aq_0\le 2$ or the denominator of the rational $bq_1\le 2$. (For example, ${x\choose 2}={x(x-1)\over 2}$ and these factors are considered "integer-valued".) But we also notice that $q_0$ must be between $0$ and the intersection $(Q_0,0)$ of the line $y=0$ with the line segment connecting $(0,p(0))$ with $\left(\frac 12,p\left(\frac 12\right)\right)$, and this distance $Q_0$ is inverse-linear with $v_3$ (by similar triangles). But this means that $a$ must be asymptotically linear with $v_3$ and also (by the same argument for $Q_1$) that $b$ must be asymptotically linear with $v_3$, and therefore that

$$p(x) = (v_3-2v_2+v_1)(x-q_0)(x-q_1)=cv_3^2(x-q_0)(x-q_1)\tag 1$$

But $v_3$ is allowed to take on any value, and equation $(1)$ implies that $c$ must vary inverse-linearly with $v_3$, therefore there exists $N\in\Bbb N$ such that for all $v_3\gt N$ at least one of $a(x-q_0),b(x-q_1)$ cannot be integer-valued.

I'm fairly certain that this same argument applies in the same way for any given value of $n$.

• Maximum possible value with regards to which parameters ? If $\nu_1, \nu_n$ are fixed, then $p$ is fixed and so is $p(n+1)$. – Joel Cohen Oct 19 '15 at 4:10
• @JoelCohen: $v_{n+1}$ is the specific parameter which determines $p(x)$, and is not otherwise restricted in any way, except that $p(x)$ is composite as described. And no, $p(x)$ is not determined except by all $n+1$ points. – abiessu Oct 19 '15 at 4:23
• Sorry, I misunderstood your question initially :) If I understand correctly, you chose the values $p(1), \ldots p(n)$ and want to know the maximal possible value of $p(n+1)$ such that $p$ is composite. Is that right ? – Joel Cohen Oct 19 '15 at 4:29
• @JoelCohen: yes that is correct. Beyond that, the $v_1,\dots,v_n$ values are given such that they do not interfere with $p(x)$ being possibly composite... – abiessu Oct 19 '15 at 4:30
• But what exactly do you mean by composite ? Real polynomials of degree $\ge 3$ are never irreducible in $\mathbb{R}[X]$ (although they might be in $\mathbb{Q}[X]$). – Joel Cohen Oct 19 '15 at 4:51

Let $p(x)$ be a given integer-valued polynomial of degree $d$. There exists a value $M$ based on $d$ consecutive integral values of $p(x)$ starting at any value $a$ (i.e., $M$ is based on $p(a),p(a+1),\dots,p(a+d-1)$) such that if $|p(a+d)|\gt M$ then $p(x)$ is irreducible over the set of integer-valued polynomials.
Proof: Let $p(x)$ be a given integer-valued polynomial of degree $d$. Suppose $p(x)$ is reducible over the set of integer-valued polynomials, having factors $q(x),r(x)$. Suppose w.l.o.g. that $d-1\ge\deg(q)\ge\deg(r)$. Consider the values $p(0), p(1), p(2),\dots,p(d-1)$. In particular, note that $q(0)\mid p(0), q(1)\mid p(1),\dots, q(d-1)\mid p(d-1)$, and this means that $q(0)$ can take on at most $\sigma_0(p(0))$ possible values, $q(1)$ can take on at most $\sigma_0(p(1))$ possible values, etc., and so there are at most
$$\prod_{i=0}^d\sigma_0(p(i))$$
possible factors $q(x)$ of $p(x)$, as these are all the possible polynomials of degree $d-1$ passing through this set of values. Each of these polynomials $q(x)$ has a value at $q(d)$, and there is a finite set of such possible values arising from the possible polynomials $q(x)$. A finite set of integers has a maximum and a minimum element, and if we take $M$ as the greatest absolute value among this minimum and maximum, we have proven the theorem.