Proof of identity for $\gcd\{f(d), d \geq 1\}$ for polynomials $f(x) \in \mathbb{Z}[x]$

Question

Reading up about Bunyakovsky conjecture, it mentions that in order to verify that polynomial $$f(x)=c_0+c_1x+\dots+c_nx^n$$ represents coprime values, or more generically in order to find $\gcd\{f(d), d \geq 1\}$, it is said to be equivalent to represent $f(x)$ in a form $$ f(x)=a_0+a_1\binom{x}{1}+\dots+a_n\binom{x}{n} $$ and find value $\gcd(a_0,a_1,\dots,a_n)=\gcd\{f(d), d \geq 1\}$ instead. Can someone reference the source of this claim, ideally a proof?

I could not find it anywhere on this site or elsewhere on the internet. All I've been able to do was to play with the coefficients and by observation find out that $$ a_i = i!\sum_{k=i}^{n}{k\brace i}c_k, $$ where ${k\brace i}$ are Stirling numbers of the second kind, although I didn't prove that either (it is probably provable by induction, but it wasn't my goal).

Answer 1

I think I've found it.

In A Survey on Fixed Divisors, page 3 mentions generalized version for multivaried polynomials (definition of $d(S,f)$ is mentioned on previous page, it is quite generic but in our case it is $\gcd$ of all polynomial values).

Thereom 2.1 (Hensel [67]) Let $f \in \mathbb{Z}[\underline{x}]$ be a polynomial with degree $m_i$ in $x_i$ for $i=1,2,\dots,n$. Then $d(\mathbb{Z}^n,f)$ is equal to the g.c.d of the values $f(r_1,r_2,\dots,r_n)$, where each $r_i$ ranges over $m_i+1$ consecutive integers.

And comments after clarifies that for $n=1$ we get desired result:

Thus, if $f(x) \in \mathbb{Z}[x]$ is a polynomial of degree $k$ then $d(\mathbb{Z},f)=(f(0),f(1),\dots,f(k))$.

So in fact, we can use any consecutive $k+1$ integer values! And the actual proof is referenced to K. Hensel, Ueber den grössten gemeinsamen Theiler aller Zahlen, welche durch eine ganze Function von n Veränderlichen darstellbar sind, J. reine angew. Math. 116 (1896), 350–356. Too bad it is in german, I am not gonna get much of it.

However, the article talks about $(f(0),f(1),\dots,f(k))$, but wiki states $(a_0,a_1,\dots,a_n)$. But later in the same article there is a Theorem 5.5 which states several equalities, this is essentially one of them.

Answer 2

Let me show that if $$f(x)=\sum_{i=0}^d\binom xia_i,$$ then $$\gcd(\{f(n):n\in\def\Z{\mathbb Z}\Z\})=\gcd(f(n_0),\dots,f(n_0+d))=\gcd(a_0,\dots,a_d)$$ for any $n_0\in\Z$.

On the one hand, it is clear that $\gcd(a_0,\dots,a_d)$ divides $f(n)$ for every $n\in\Z$, as each $\binom ni$ is an integer. I will prove the other direction $$\forall n_0\in\Z\colon\quad\gcd(f(n_0),\dots,f(n_0+d))\mid a_0,\dots,a_d$$ by induction on $d$. For $d=0$, $f(x)=a_0$, hence the result is clear.

Assume the statement holds for $d-1\ge0$; we will prove it for $d$. Putting $g=\gcd(f(n_0),\dots,f(n_0+d))$, the difference polynomial $$\Delta f(x)=f(x+1)-f(x)=\sum_{i<d}\binom xia_{i+1}$$ satisfies $$g\mid \Delta f(n_0),\dots,\Delta f(n_0+d-1),$$ hence by the induction hypothesis, $$g\mid a_1,\dots,a_d.$$ Thus, $g$ divides $$a_0=f(n_0)-\sum_{i=1}^d\binom{n_0}ia_i$$ as well.