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Math Help - Perfect square

  1. #1
    Super Member dhiab's Avatar
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    Perfect square

    Find all naturel numbers n such as : n^4 - 4n^3 + 22n^2 - 36n + 18 is the Perfect square .
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  2. #2
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    Hello, dhiab!

    Is there a typo?
    If the last constant is 81, there is a simple solution.


    Find all natural numbers n such that: . n^4 - 4n^3 + 22n^2 - 36n + {\color{red}81} . is a perfect square.

    Let: . (n^2 + an + b)^2 \:=\:n^4 - 4n^3 + 22n^2 - 36n + 81

    Expand: . n^4 + 2an^3 + (a^2+2b)n^2 + 2abn + b^2 \;=\;n^4 - 4n^3 + 22n^2 - 36n + 81


    And we have: . \begin{array}{c} 2a \:=\:-4 \\ a^2+2b \:=\:22 \\ 2ab \:=\:-36 \\ b^2 \:=\:81 \end{array} \quad\Rightarrow \quad a \:=\:-2,\;b \:=\:9


    Therefore: . (n^2 - 2n + 9)^2 \;=\;n^4 - 4n^3 + 22n^2 - 37n + 81

    The expression is a square for all natural numbers n.

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