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).

  • 1
    $\begingroup$ This answer has a proof of this claim. $\endgroup$
    – robjohn
    Mar 17, 2019 at 2:12

1 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.


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