Results 1 to 4 of 4

Thread: Quickie #8

  1. #1
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    849

    Quickie #8

    Prove that for any positive integer $\displaystyle n$

    . . $\displaystyle n^4 + 2n^3 + 2n^2 + 2n + 1$ is never a perfect square.

    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by Soroban View Post
    Prove that for any positive integer $\displaystyle n$

    . . $\displaystyle n^4 + 2n^3 + 2n^2 + 2n + 1$ is never a perfect square.

    $\displaystyle (n^4+2n^2+1)+(2n^3+2n)$
    $\displaystyle (n^2+1)^2+2n(n^2+1)$
    $\displaystyle (n^2+1)((n^2+1)+2n)$
    $\displaystyle (n^2+1)(n^2+2n+1)$
    $\displaystyle (n^2+1)(n+1)^2$
    Assume it is a square....
    Theorem: If a product of a positive integer and a square is is square then that positive integer itself must be a square.

    Thus,
    $\displaystyle n^2+1=m^2$
    $\displaystyle m^2-n^2=1$
    $\displaystyle (m+n)(m-n)=1$
    Trivial-Improper factorization,
    $\displaystyle m+n=1$
    $\displaystyle m-n=1$
    Thus,
    $\displaystyle n=0$.
    But the problem says positive integer.... a contradiction.

    This is mine 4th Post!!!
    Last edited by ThePerfectHacker; Jan 6th 2007 at 02:43 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,002
    Thanks
    1
    Here's a thought:

    If you factor:

    $\displaystyle (n+1)^{2}(n^{2}+1)=k^{2}$

    Take the square root:

    $\displaystyle (n+1)\sqrt{n^{2}+1}=k$

    If n is a positive integer, then k can not be an integer, therefore, can not be a perfect square.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    849
    Hello, Hacker and Galactus!

    I like both your solutions.

    The "Quickie" solution is surprisingly clever
    . . but almost impossible to derive.


    Let $\displaystyle N \:=\:n^4 + 2n^3 + 2n^2 + 2n + 1$

    Since $\displaystyle n^4 + 2n^3 + n^2 \:< \:n^4 + 2n^3 + 2n^2 + 2n + 1$
    . . we have: .$\displaystyle (n^2 + n)^2 \:< \:N$

    Since $\displaystyle n^4 + 2n^3 + 2n^2 + 2n + 1 \:< \:n^4 + 2n^3 + 3n^2 + 2n + 1$
    . . we have: .$\displaystyle N \:< \:(n^2 + n + 1)^2$


    Hence: .$\displaystyle (n^2 + n)^2 \:< \:N \:< \:(n^2 + n + 1)^2$
    . . and $\displaystyle N$ lies between two consecutive squares.

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Quickie #15
    Posted in the Math Challenge Problems Forum
    Replies: 3
    Last Post: Jan 25th 2007, 01:46 PM
  2. Quickie #14
    Posted in the Math Challenge Problems Forum
    Replies: 3
    Last Post: Jan 20th 2007, 07:00 AM
  3. Quickie #13
    Posted in the Math Challenge Problems Forum
    Replies: 7
    Last Post: Jan 18th 2007, 06:20 PM
  4. Quickie #12
    Posted in the Math Challenge Problems Forum
    Replies: 4
    Last Post: Jan 16th 2007, 11:25 AM
  5. Quickie #1
    Posted in the Math Challenge Problems Forum
    Replies: 2
    Last Post: Dec 23rd 2006, 06:02 AM

Search Tags


/mathhelpforum @mathhelpforum