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Math Help - Pythagorean Triples II

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    Pythagorean Triples II

    Find all the Pythagorean triples that form an arithmetic sequence.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by MATNTRNG View Post
    Find all the Pythagorean triples that form an arithmetic sequence.
    Is this a challenge problems? If not, let's see some work.
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    Hello, MATNTRNG!

    Find all the Pythagorean triples that form an arithmetic sequence.

    Assume that Pythagorean triples consist of positive integers.


    We have: . (a,\:a+d,\:a+2d)

    Hence: . a^2 + (a+d)^2 \:=\:(a+2d)^2

    . . a^2 + a^2 + 2ad + d^2 \:=\:a^2 + 4ad + 4d^2

    . . . . a^2 -2ad - 3d^2 \:=\:0

    . . . . (a+d)(a-3d) \:=\:0

    And we have: . \begin{Bmatrix}a+d &=& 0 & \Rightarrow & d &=& \text{-}a \\<br />
a - 3d &=& 0 & \Rightarrow & d &=& \frac{a}{3} \end{Bmatrix}


    Since d = \frac{a}{3}, the Pythagorean triples are: . \left(a,\:\frac{4}{3}a,\:\frac{5}{3}a\right)
    . . where \,a is any positive multiple of 3.
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    A useful strategy would be to denote the sides a-d, a, and a+d, where a and b are integers; these form an arithmetic sequence. We assume that d>0.

    The following relation holds: (a-d)^2+a^2=(a+d)^2. From a^2-2ad+d^2+a^2=a^2+2ad+d^2, we find that a^2=4ad. But a>0 since a-d is one of the triangle's sides, so we can divide by a to get a=4d.

    Thus Pythagorean triples that form an arithmetic sequence are given by 3d, 4d, 5d. In particular, this implies that the only Pythagorean triple that consists of consecutive postive integers is 3, 4, 5.
    Last edited by melese; November 2nd 2010 at 08:20 AM. Reason: additional information
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