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Math Help - In the right triangles with sides...

  1. #1
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    In the right triangles with sides...

    In the right triangles with sides (3,4,5) and (5,12,13), one side is just 1 less than the hypotenuse. Find all such integer-sided right triangles. [If you cannot do that, at least try to determine whether or not there are infinitely many of them.]

    Thanks!
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by NikoBellic View Post
    In the right triangles with sides (3,4,5) and (5,12,13), one side is just 1 less than the hypotenuse. Find all such integer-sided right triangles. [If you cannot do that, at least try to determine whether or not there are infinitely many of them.]

    Thanks!
    Since  c is odd and  c-b=1 , this implies that  b is even, so  b=2mn and we know  c=m^2+n^2 .

    So  c-b = m^2-2mn+n^2 = (m-n)^2=1\implies m=n+1 .

    Thus  a=m^2-n^2 = (n+1)^2-n^2 = 2n+1

     b=2mn = 2(n+1)n = 2n^2+2n

     c=m^2+n^2 = (n+1)^2+n^2 = 2n^2+2n+1 where  n is arbitrary.
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    Hello, NikoBellic!

    In the right triangles with sides (3,4,5) and (5,12,13),
    . . one side is just 1 less than the hypotenuse.
    Find all such integer-sided right triangles.

    Pythagorean triples are generated by: . \begin{Bmatrix}a &=& m^2-n^2 \\ b &=& 2mn \\ c &=& m^2+n^2 \end{Bmatrix} . where: . m,n \in I^+\:\text{ and }\,m > n.


    Let m \:=\:n+1 \qquad\text{Then: }\:\begin{Bmatrix}a &=& 2n+1 \\ b &=& 2n^2+2n \\ c &=& 2n^2+2n+1 \end{Bmatrix}


    And we have: . (3,4,5),\;(5,12,13),\;(7,24,25),\;(9,40,41),\;(11,  60,61),\;\hdots


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