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Math Help - Write as whole square

  1. #1
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    Write as whole square

    Can you please write equation
    n(n+1)(n+2)(n+3)+1
    as whole square

    like writing {x}^2+4x+4=(x+2)(x+2) = {(x+2)}^{2}

    My Try:
    n(n+1)(n+2)(n+3)+1={n}^{4}+{6n}^{3}+{11n}^{2}+6n+1
    and this equation has 2 pairs of equal roots, and they are not integers
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  2. #2
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    Quote Originally Posted by razemsoft21 View Post
    Can you please write equation
    n(n+1)(n+2)(n+3)+1
    as whole square

    like writing {x}^2+4x+4=(x+2)(x+2) = {(x+2)}^{2}

    My Try:
    n(n+1)(n+2)(n+3)+1={n}^{4}+{6n}^{3}+{11n}^{2}+6n+1
    and this equation has 2 pairs of equal roots, and they are not integers
    factorise x(x+1)(x+2)(x+3&#4 1; + 1 - Wolfram|Alpha
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  3. #3
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    Correct answer thank you.
    can you please show me the way you solve it ?
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  4. #4
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    Quote Originally Posted by razemsoft21 View Post
    Correct answer thank you.
    can you please show me the way you solve it ?
    Can you reverse engineer a solution from the answer ...? See how you go.
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  5. #5
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    Hello, razemsoft21!

    Write as a square: . n(n+1)(n+2)(n+3)+1

    We have: . \bigg[n(n+3)\bigg]\,\bigg[(n+1)(n+2)\bigg] + 1

    . . =\;\bigg[n^2 + 3n\bigg]\,\bigg[n^2+3n + 2\bigg] +1

    . . =\; \bigg[n^2 + 3n +1 - 1\bigg]\,\bigg[n^2 + 3n + 1 + 1\bigg] + 1

    . . =\; \bigg[(n^2 + 3n +1) - 1\bigg]\,\bigg[(n^2 + 3n + 1) + 1\bigg] + 1

    . . =\;\bigg[(n^2+3n + 1)^2 - 1^2\bigg] + 1

    . . =\;(n^2 + 3n + 1)^2
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