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Pythagorean Triangles with one side equal s and hypothenuse equal 2 s+1

  1. Aug 13, 2011 #1
    This thread is concerned with rectangular triangle with one side ( = s) and hypothenuse
    (t = 2 s + 1) given and we are looking for the remaining side (all sides having integer values)

    d[itex]^{2}[/itex] = t[itex]^{2}[/itex] - s[itex]^{2}[/itex] = (2 s + 1)[itex]^{2}[/itex] - s[itex]^{2}[/itex], with d, s, t [itex]\in \mathbb{Z}[/itex]

    Computational experiments show the first solutions to this problem as:

    s = 0 | 8 | 120 | 1680 | 23408 | ...
    t = 1 | 17 | 241 | 3361 | 46817 | ...
    d = 1 | 15 | 209 | 2911 | 40545 | ...

    Further mathematical treatment starts with:

    Lemma: Let b[itex]_{0}[/itex] and b[itex]_{1}[/itex] be co-prime natural numbers and

    b[itex]_{k}[/itex]:=4 b[itex]_{k-1}[/itex] - b[itex]_{k-2}[/itex]

    then the pairs (b[itex]_{k},b_{k+1}[/itex]) are all co-prime

    Euclidean Rule for Pythagorean Numbers:
    Let (m,n) be co-prime natural numbers (m<n), then

    h := n[itex]^{2}[/itex] + m[itex]^{2}[/itex]
    e := 2 m n
    d := n[itex]^{2}[/itex] - m[itex]^{2}[/itex]

    form the hypothenuse, the even and the odd leg
    of a primitive Pythagorean triangle (PPT)


    Now we have from b = {0, 1, 4, 15, 56, 209, 241, ...}

    (m.n) -h- -e- -d-
    ---------------------------------
    (0,1) -1- -0- -1-
    (1,4) -17- -8- -15-
    (4,15) -241- -120- -209-
    (15,56) -3361- -1680- -2911-
    (56,209) -46827- -23408- -40545-
    (...,...)

    and we identify the PPT's generated from the b-sequence as solutions to our problem.
     
  2. jcsd
  3. Aug 13, 2011 #2
    Further analysis in this topic will be much easier
    with the following closed form equation for b[itex]_{k}[/itex]


    b[itex]_{k}[/itex] := [itex]\frac{(2+\sqrt{3})^{k} - (2-\sqrt{3})^{k}}{2 \sqrt{3}}[/itex]
     
  4. Aug 13, 2011 #3
    With the equation for b[itex]_{k}[/itex] in closed form as above and the definitions
    for then even resp odd leg and the hypothenuse of the Euclidean PTT rule,
    we have:

    d[itex]_{k} := \frac{1}{6} ((3+2 \sqrt{3}) (2+\sqrt{3})^{2 k} + (3-2 \sqrt{3}) (2-\sqrt{3})^{2 k})[/itex]

    e[itex]_{k}:=\frac{1}{6} ((2+\sqrt{3})^{(1+2 k)}+(2-\sqrt{3})^{(1+2 k)} - 4)[/itex]

    h[itex]_{k} := \frac{1}{3} ((2+\sqrt{3})^{(1+2 k)}+(2-\sqrt{3})^{(1+2 k)} - 1)[/itex]

    and from the last two equations, it is easy to see, that

    h[itex]_{k} = 2 e_{k}[/itex] + 1
     
  5. Aug 15, 2011 #4
    Let b[itex]_{0}[/itex]=0, b[itex]_{1}[/itex]=1 and b[itex]_{k}[/itex]:=4 b[itex]_{k-1}[/itex] - b[itex]_{k-2}[/itex] as above,
    we have for the odd leg of aa PTT after the Euclidean rule for PTT's

    d[itex]_{k}[/itex]= b[itex]_{k+1}[/itex][itex]^{2}[/itex] - b[itex]_{k}[/itex][itex]^{2}[/itex]

    Numerical inspection of the so defined sequences show:

    b = {0,1,4,15,56,209,780,2911,10864,40545,...} and

    d = {15,209,2911,40545, ...}

    the highly remarkable fact d[itex]_{k}[/itex]= b[itex]_{2k+1}[/itex]
     
  6. Aug 20, 2011 #5
    Related to our problem is:

    Pythagorean Triangles with one side equal s and hypothenuse equal 2 s-1

    Computational experiments of the first solutions (given are the even leg e, the odd leg d
    and the hypothenuse h, with the corresponding (m,n) from the Êuclidean Rule for PPT):

    i - (m.n) - d / e / h
    -------------------
    1 - (1,2) - 3 / 4 / 5
    2 - (4,7) - 33 / 56 / 65
    3 - (15,26) - 451 / 780 / 901
    4 - (56,97) - 6273 / 10864 / 12545
    5 - (209,362) - 87363 / 151316 / 174725

    We see that the m[itex]_{k}[/itex] and the n[itex]_{k}[/itex] form the recurrence relation:

    m[itex]_{k}[/itex] = 4 m[itex]_{k-1}[/itex] - m[itex]_{k-2}[/itex] and

    n[itex]_{k}[/itex] = 4 n[itex]_{k-1}[/itex] - n[itex]_{k-2}[/itex]

    with different starting values.

    That gives hope for an interesting investigation in the problem:

    Pythagorean Triangles with one side equal s and hypothenuse equal 2 s+k, k[itex]\in \mathbb{Z}[/itex]
     
  7. Sep 23, 2011 #6
    related paper:

    http://conservancy.umn.edu/bitstream/4878/1/438.pdf [Broken]

    and some of my ideas:

    http://dl.dropbox.com/u/13155084/2D/Fourier.html [Broken]

    http://dl.dropbox.com/u/13155084/Pythagorean%20lattice.pdf [Broken]
     
    Last edited by a moderator: May 5, 2017
  8. Oct 15, 2011 #7
    Let k = 3a^2-b^2 then the series for m begins {a,2a+b,7a+4b,...} and the series for n begins {b,3a+2b,12a+7b,...}. This does not preclude other values of m and n from also solving the relationship one side = s and the hypothenuse equals 2s + k though. Example, m^2+n^2 - 2(n^2-m^2) = 3m^2-n^2 = 3*(2a+b)^2 - (3a+2b)^2 = 3*(7a+4b)^2 - (12a-7b)^2. As one can see, the relation m_n = 4 m_(n-1) - m_(n-2) holds for this pattern. Since -13 = 3*1^2-4^2 = 3*2^2 -5^2 one can see how different sequences may hold for the same value of k. Still working on other relationships e.g. for the twice the even side + k = the hypothenuse.
     
  9. Oct 15, 2011 #8
    For the hypothenuse - twice the even side we have a^2 + b^2 -4ab = b^2 + (4b-a)^2 - 4b*(4b-a). In other words m_0 = a, m_1 = b m_2 = 4b-a and n_0 = b, n_1 = 4b-a, n_2 = 15b-4a. Both have the recurrence relation S(n) = 4*S(n-1) - S(n-2)

    For the even side -the odd side, 4ab - b^2 + a^a = 4*(4b+a)*(17b+4a)-(17b+4a)^2 + (4b+a)^2. In other words, m_0 = a, m_1 = a+4b, m_2 = 17a + 72b and n_0 = b, n_1 = 17b + 4a, n_2 = 305b + 72a. Both have the recurrence relation S_n = 18*S(n-1)-S(n-2) .
     
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