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Math Help - Working with triangular numbers

  1. #1
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    Working with triangular numbers

    The n−th triangular number is given by T_{n} = \frac{1}{2}n(n+1). how can one find all integers n for which the sum of the (n − 1)−st, the n−th,
    and the (n + 1)−st triangular numbers is a square?
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  2. #2
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    Quote Originally Posted by Intsecxtanx View Post
    The n−th triangular number is given by T_{n} = \frac{1}{2}n(n+1). how can one find all integers n for which the sum of the (n − 1)−st, the n−th,
    and the (n + 1)−st triangular numbers is a square?
    Google told me to look in the papers:
    Triangular Numbers and Perfect Squares
    Tom Beldon and Tony Gardiner
    The Mathematical Gazette
    Vol. 86, No. 507 (Nov., 2002), pp. 423-431
    JSTOR: An Error Occurred Setting Your User Cookie

    88.53 Comments on 'Triangular Numbers and Perfect Squares'
    Michael D. Hirschhorn
    The Mathematical Gazette
    Year, volume, issue, and the page range: Vol. 88, No. 513 (Nov., 2004), pp. 500-503
    JSTOR: An Error Occurred Setting Your User Cookie

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  3. #3
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Intsecxtanx View Post
    The n−th triangular number is given by T_{n} = \frac{1}{2}n(n+1). how can one find all integers n for which the sum of the (n − 1)−st, the n−th,
    and the (n + 1)−st triangular numbers is a square?
     T_{n-1}+T_n+T_{n+1} = \frac12(3n^2+3n+2)

    We need to solve the Diophantine equation  3n^2+3n+2 = 2m^2 . Let's see if you can solve this before I dive right in.
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  4. #4
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    Diophantine equations are my weakness. I know how to tackle pell's equations with "continued fractions" but usually I don't know how else to solve them with out random guess and check methods.
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  5. #5
    MHF Contributor chiph588@'s Avatar
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    We have  3n^2+3n+2=2m^2

    Now multiply by  12 and add and subtract 9 to get  (6n+3)^2-24m^2+15=0

    Let  y=6n+3 and multiply both sides by  24 to get  (24m)^2-24y^2=360

    By letting  x=24m we see we have  x^2-24y^2=360 which is almost Pell's Equation.

    Luckily though this is solved similarly. I won't solve it, but here for reference on how to solve it (look for the part about solving  x^2-ny^2=c ).
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