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Thread: Pythagorean triple in (a,a+1,c)

  1. #1
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    Pythagorean triple in (a,a+1,c)

    Pythagorean triple is (a,b,c) satisfies $\displaystyle a^2+b^2=c^2$

    Find all Pythagorean triple in (a,a+1,c)

    the first two are (3,4,5) and (20,21,29)

    thanks!!!
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  2. #2
    Super Member PaulRS's Avatar
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    We require: $\displaystyle a^2+(a+1)^2=c^2$

    Expand: $\displaystyle 2a^2+2a+1=c^2$

    Now multiply by 2: $\displaystyle 4a^2+4a+2=2c^2$

    So: $\displaystyle 4a^2+4a+1+1=2c^2$ and note that: $\displaystyle 4a^2+4a+1=(2a+1)^2$

    We have: $\displaystyle
    \left( {2a + 1} \right)^2 + 1= 2c^2
    $

    Set $\displaystyle x=2a+1$ and $\displaystyle y=c$

    So our condition turns into: $\displaystyle x^2+1=2y^2$

    And this is a Pell Type equation ! (Read here, if you have any doubts, ask )
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  3. #3
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    Hello, chipai!

    Pythagorean triple is $\displaystyle (a,b,c)$ satisfies $\displaystyle a^2+b^2\:=\:c^2$

    Find all Pythagorean triple of the form: $\displaystyle (a,a+1,c)$

    The first two are $\displaystyle (3,4,5)$ and $\displaystyle (20,21,29)$.
    The first few are:

    . . $\displaystyle \begin{array}{ccc} a & b & c \\ \hline
    3 & 4 & 5 \\ 20 & 21& 29 \\ 119& 120& 169 \\ 696 & 697 & 985 \\ 4059 & 4060 & 5741 \end{array}$


    The recursive formulas are:

    . . $\displaystyle \begin{array}{ccc}a_n &=& 6a_{n-1} - a_{n-2} + 2 \\ \\[-3mm]
    c_n &=& 6c_{n-1} - c_{n-2}\qquad \end{array} $

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