Results 1 to 4 of 4

Thread: Not eventually constant!

  1. #1
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7

    Not eventually constant!

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

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


    Remark: This problem is an easy case of a long standing conjecture in number theory: if $\displaystyle g(x) \in \mathbb{Z}[x]$ and $\displaystyle f(g(n))$ is "eventually constant", then $\displaystyle g(x)$ 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

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

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

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

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


    Last edited by Bruno J.; Oct 23rd 2009 at 08:32 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

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


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

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

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


    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: Mar 26th 2011, 01:42 PM
  2. question about eventually decreasing sequences
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: Sep 23rd 2010, 12:58 PM
  3. Eventually bounded rational functions
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Apr 11th 2010, 11:26 AM
  4. sequences of eventually monotone functions
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: Sep 28th 2009, 09:59 PM
  5. Replies: 5
    Last Post: Feb 29th 2008, 02:05 PM

Search Tags


/mathhelpforum @mathhelpforum