TITLE: How does one show that two general numbers $n! + 1$ and $(n+1)! + 1$ are relatively prime?
QUESTION [19 upvotes]: How does one show that two general numbers $n! + 1$ and $(n+1)! + 1$ are relatively prime?
I don't mind if someone uses a different example, I want to learn how to prove this class of problems.
My professor seems to really emphasize that you can use the division algorithm, or find the greatest common divisor using the division algorithm rather, in order to assess the nature of relative-prime-ness and I am kind of confused as to how to apply it.

REPLY [5 votes]: Hint $\ $ Put $\, k = n!\,$ in the following Euclidean gcd reduction
$$(k\!+\!1,(n\!+\!1)k\!+\!1) = (\color{#c00}{k\!+\!1},(n\!+\!1)(\color{#c00}{k\!+\!1})\!-\!n) = (\color{}{k\!+\!1},-n)\,\ \ \left[\,= 1\ \ {\rm if}\ \ n\mid k\,\right]$$
Generally $\ (k\!+\!1,f(k)) = (k\!+\!1,f(-1))\,\ $ [$=1\,$ if $\,f(-1)\mid k\,$] where the first equality employs gcd modular reduction combined with the polynomial remainder theorem, i.e. $\bmod k\!+\!1\!:\ \,\color{#c00}{k\equiv -1}\Rightarrow f(\color{#c00}k)\equiv f(\color{#c00}{-1}),\,$ for any polynomial $\,f(x)\,$ with integer coefficients. Above we have $\,f(x) = (n\!+\!1)x+1\,$ so $\,f(-1) = -n$