# How to find minimum n that $\gcd(a+n^b, a+(n+1)^b) \neq 1$?

After reading an answer of a problem, saying that $\gcd(9+n^{17},9+(n+1)^{17}) = 1$ holds until n=8424432925592889329288197322308900672459420460792433, my first though was 'How the heck they found it?.'

So my question is

How to calculate the minimum $n \in \mathbb{N}$ that $\gcd(a+n^b , a+(n+1)^b) \neq 1 (a,b \in \mathbb{N})$, if it exists?

I think mathematicians didn't blindly brute-forced it, because 84...33 is a very large number, so any (super)computer couldn't do that.

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Let $f(x)=x^{17}+9$, $g(x)=(x+1)^{17}+9$. Suppose $\gcd(f(n),g(n))=d\ne1$. Let $p$ be a prime dividing $d$. Then $f(x)$ and $g(x)$ have a common zero (namely, $x=n$) when viewed as polynomials over the field of $p$ elements. Two polynomials having a common zero is the same as their resultant being zero; having a common zero over the field of $p$ elements is the same as the resultant being divisible by $p$. So you compute the resultant of $f$ and $g$, and look for a prime factor.