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Math Help - Quickie #8

  1. #1
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    Quickie #8

    Prove that for any positive integer n

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

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  2. #2
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    Quote Originally Posted by Soroban View Post
    Prove that for any positive integer n

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

    (n^4+2n^2+1)+(2n^3+2n)
    (n^2+1)^2+2n(n^2+1)
    (n^2+1)((n^2+1)+2n)
    (n^2+1)(n^2+2n+1)
    (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,
    n^2+1=m^2
    m^2-n^2=1
    (m+n)(m-n)=1
    Trivial-Improper factorization,
    m+n=1
    m-n=1
    Thus,
    n=0.
    But the problem says positive integer.... a contradiction.

    This is mine 4th Post!!!
    Last edited by ThePerfectHacker; January 6th 2007 at 02:43 PM.
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  3. #3
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    Here's a thought:

    If you factor:

    (n+1)^{2}(n^{2}+1)=k^{2}

    Take the square root:

    (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.
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  4. #4
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    Hello, Hacker and Galactus!

    I like both your solutions.

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


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

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

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


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

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