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Math Help - finding all squares

  1. #1
    Member Veronica1999's Avatar
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    finding all squares

    Could I get some help on this problem?

    Thanks.
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Re: finding all squares

    Let n^2+9n+14=a^2, where a is an integer.

    n^2+9n+14-a^2=0

    Let us now calculate the discriminant of this quadratic equation in n.

    \Delta=81-4(14-a^2)=25+4a^2=5^2+(2a)^2

    In order for n to be an integer, we need \Delta to be a perfect square.

    The only Pythagorean triple that has 5 as one of the smaller values is 5, 12, 13.

    So, a=\pm 6

    n^2+9n+14=36

    n^2+9n-22=0

    (n+11)(n-2)=0

    n=-11,\ 2
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