1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Prove gcd(nn!, n!+1)=1

  1. Apr 10, 2012 #1
    1. The problem statement, all variables and given/known data
    For any [itex]n \in \mathbb{N}[/itex], find [itex]\mathrm{gcd}(n!+1,(n+1)!+1)[/itex]. First come up with a conjecture, then prove it.

    2. The attempt at a solution
    By testing some values, it seems like [itex]\mathrm{gcd}(n!+1,(n+1)!+1) = 1[/itex]

    I'm trying to prove this by induction. I'll leave out the inductive assumption and base case verification because I can do those.

    I have [itex]\mathrm{gcd}(n!+1,(n+1)!+1) = 1[/itex] and I'm trying to show that [itex]\mathrm{gcd}((n+1)!+1,(n+2)!+1) = 1[/itex].

    I can simplify what's given to me to [itex]\mathrm{gcd}(nn!, n!+1)=1[/itex] but I can't find out how to get it into the form I want it. Can anybody look at what I'm doing and give me any guidance?

    [itex]\mathrm{gcd}(n!+1,(n+1)!+1) = 1 \implies \mathrm{gcd}(n!+1,(n+1)n!+1) = 1 \implies \mathrm{gcd}(n!+1,nn!+n!+1) = 1 \implies \mathrm{gcd}((n)n!, n!+1) = 1[/itex]
     
  2. jcsd
  3. Apr 11, 2012 #2

    morphism

    User Avatar
    Science Advisor
    Homework Helper

    I wouldn't bother with induction here.

    Try to argue as follows. If a prime divides both n!+1 and (n+1)!+1, it will divide their difference. What does this imply?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook