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Thread: Working with triangular numbers

  1. #1
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    Working with triangular numbers

    The n−th triangular number is given by $\displaystyle T_{n} = \frac{1}{2}n(n+1)$. 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 $\displaystyle T_{n} = \frac{1}{2}n(n+1)$. 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 $\displaystyle T_{n} = \frac{1}{2}n(n+1)$. 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?
    $\displaystyle T_{n-1}+T_n+T_{n+1} = \frac12(3n^2+3n+2) $

    We need to solve the Diophantine equation $\displaystyle 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 $\displaystyle 3n^2+3n+2=2m^2 $

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

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

    By letting $\displaystyle x=24m $ we see we have $\displaystyle 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 $\displaystyle x^2-ny^2=c $).
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