Results 1 to 4 of 4

Math Help - Not eventually constant!

  1. #1
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7

    Not eventually constant!

    Let f: \mathbb{N}: \longrightarrow \{1,-1 \} be a function such that f(p)=-1, for all prime numbers p, and f(mn)=f(m)f(n), \ \forall m, n \in \mathbb{N}.

    Prove that for every k \in \mathbb{N} there exist m, n \geq k such that f(n^2-2)=1 and f(m^2 - 2)=-1.


    Remark: This problem is an easy case of a long standing conjecture in number theory: if g(x) \in \mathbb{Z}[x] and f(g(n)) is "eventually constant", then g(x) is a perfect square.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1
    We have the identity

    (n^2-2)\left((n+1)^2-2\right)=\left((n^2+n-2)^2-2\right)

    which shows that it's impossible for the sequence f(n^2-2) to be eventually constantly equal to -1. If it is eventually constantly equal to 1, let s be the greatest integer such that f(s^2-2)=-1. Then by hypothesis f((s+1)^2-2)=1 and then -1 = f(s^2-2)f((s+1)^2-2)=f\left((s^2+s-2)^2-2\right), contradicting the choice of s.

    On a side note, I believe f(n) is called Liouville's function.

    You always have nice problems, NonCommLion. Where do you get them from?


    Last edited by Bruno J.; October 23rd 2009 at 09:32 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by Bruno J. View Post
    We have the identity

    (n^2-2)\left((n+1)^2-2\right)=(n^2+n-2)^2-2


    which shows that it's impossible for the sequence f(n^2-2) to be eventually constantly equal to -1. If it is eventually constantly equal to 1, let s be the greatest integer such that f(s^2-2)=-1. Then by hypothesis f((s+1)^2-2)=1 and then -1 = f(s^2-2)f((s+1)^2-2)=f\left((s^2+s-2)^2-2\right), contradicting the choice of s.
    nice! just note that before choosing the greatest integer s with f(s^2-2)=-1 you should mention that basically the set A=\{n: \ f(n^2-2)=-1 \} is non-empty.

    this, of course, is trivial in our case because 2 \in A, but it becomes absolutely non-trivial if we replace -2 with an arbitrary non-zero integer k.

    finally the identity you used can be extended. in general we have: (n^2+k)((n+1)^2+k)=(n^2+n+k)^2+k.


    You always have nice problems, NonCommLion.
    thanks! i like my new user name: NCL! it could also stand for Normal CLosure!


    Where do you get them from?
    here and there!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1
    Quote Originally Posted by NonCommAlg View Post

    thanks! i like my new user name: NCL! it could also stand for Normal CLosure!


    here and there!
    Well in any case keep them coming! I can't do most of them but they are fun.

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: March 26th 2011, 02:42 PM
  2. question about eventually decreasing sequences
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: September 23rd 2010, 01:58 PM
  3. Eventually bounded rational functions
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: April 11th 2010, 12:26 PM
  4. sequences of eventually monotone functions
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: September 28th 2009, 10:59 PM
  5. Replies: 5
    Last Post: February 29th 2008, 03:05 PM

Search Tags


/mathhelpforum @mathhelpforum