Smallest Right Triangle

Discussion in 'Recreational Math' started by Jim Dars, Dec 16, 2006.

  1. Jim Dars

    Jim Dars Guest

    Hi All,

    Consider a right triangle whose legs, X and Y, and hypotenuse Z are all
    integers.
    What is the smallest such triangle such that

    1) X + Y is a square
    2) Z is a square

    Not an easy problem. Merry Christmas and

    Best wishes, Jim
     
    Jim Dars, Dec 16, 2006
    #1
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  2. Jim Dars

    Proginoskes Guest

    Not too hard, though, once you know the formula that generates all
    Pythagorean triples:

    X = u^2 - v^2
    Y = 2 u v
    Z = u^2 + v^2,

    where u > v > 0. (If you insist further that gcd(u,v) = 1, you get all
    "primitive" Pythagorean triples:
    gcd (X,Y,Z) = 1.)

    Problem (2) is easily answered; Z >= 3^2 + 4^2 = 5^2. Thus u = 4, v =
    3, which gives
    (X,Y,Z) = (9, 24, 25).

    To answer problem (1), you need to find the smallest u and v such that
    u^2 + 2 u v - v^2 is a perfect square. If we let this expression equal
    K^2, then
    2 u^2 - K^2 must be a perfect square. From a Maple search, it appears
    that
    u = 5 and v = 4 (and hence (X,Y,Z) = (9, 40, 41)) is the solution to
    (1).

    --- Christopher Heckman
     
    Proginoskes, Dec 17, 2006
    #2
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  3. Yah v.good. Now solve it again as if it were one problem with two constraints.
     
    Patrick Hamlyn, Dec 17, 2006
    #3
  4. Jim Dars

    Jim Dars Guest

    Hi Proginoskes,

    Not even close. You must understand I posed ONE problem, not two. So X + Y
    is a square for the same triangle Z is a square.

    Best wishes, Jim
     
    Jim Dars, Dec 17, 2006
    #4
  5. Jim Dars

    Michael Guest

    Jim Dars, not easy? You can say that again!
    This is a problem for the big boys
    (Fermat, Euler, Lagrange, etc.).
    Fermat first proposed and solved it.
    Smallest triplet has sides which are 13-digit integers.
    --Michael
     
    Michael, Dec 17, 2006
    #5
  6. Jim Dars

    Proginoskes Guest

    Oh. Well, usually, that's indicated by putting the word AND between the
    two conditions. Based on other responses to this thread, that will take
    a little more time.

    --- Christopher Heckman
     
    Proginoskes, Dec 17, 2006
    #6
  7. Jim Dars

    Chip Eastham Guest

    Extending Christopher's approach to Z being a square as well
    as X+Y, we express u = m^2 - n^2 and v = 2mn, & then we ask
    X+Y = u^2 + 2uv - v^2 = m^4 + 4m^3n - 6m^2n^2 - 4mn^3 + n^4
    to be a square. Setting r = m/n, we can then search for the
    rational points on:

    y^2 = r^4 + 4r^3 - 6r^2 - 4r + 1 = f(r)

    with the proviso that X,Y > 0, which wlog implies u > v > 0:

    m^2 - n^2 > 2mn > 0 (and wlog m,n > 0)
    (m - n)^2 = m^2 - 2mn + n^2 > 2n^2
    m - n > n*sqrt(2)
    m > n*(1 + sqrt(2))
    r > 1 + sqrt(2) ~ 2.41421...

    Note that the curve above has rational points at r = 0, 3/2.
    More than likely this makes it tractable to compute more
    rational points on the elliptic(?) curve.

    regards, chip
     
    Chip Eastham, Dec 18, 2006
    #7
  8. Jim Dars

    Jim Dars Guest

    Hi Chip, Michael, All,

    Michael named Fermat quite correctly. Least I give my friends a Black
    Holiday present I'll note that he used the process of "Infinite Descent",
    but then goes in the reverse direction. Not much of a clue, and it wasn't
    meant to be. --- Just concerned that people don't waste time going down
    blind (or difficult) alleys.

    Best wishes, Paul
     
    Jim Dars, Dec 18, 2006
    #8
  9. Jim Dars

    Michael Guest

    Chip, I suspect that you may be on the right track or on a right track. A
    successive search for rational points by the "Method of Diaphantus" (in
    ascent or descent, as used by Fermat, as stated by Jim Dars) *may* be the
    trick or a trick.

    regards, michael
     
    Michael, Dec 18, 2006
    #9
  10. Jim Dars

    a1jrj Guest

    I gave up quickly with this, although the solution found by Fermat as
    noted uses fairly elementary stuff, as in this link...
    <http://www.mathpages.com/home/kmath022/kmath022.htm>
    HTH
    JJ
     
    a1jrj, Dec 18, 2006
    #10
  11. Jim Dars

    Michael Guest



    Christopher and Chip, there is nothing new below.
    I am restating your work, dotting the i's and crossing the t's, mainly for
    my own sake.

    Given:
    (X,Y,Z,W) are positive integers; gcd(X,Y) = 1
    X^2 + Y^2 = Z^2 (1)
    Z= W^2 (2)
    X + Y = W^2 (3)

    Find smallest (X,Y)

    Let (u,v) be positive integers, one odd, one even; u > v ; gcd(u,v) = 1 :
    X = (u^2 - v^2) (odd)
    Y = 2uv (even)
    Z = (u^2 + v^2) (odd)
    Then (X,Y,Z) is a solution of (1).

    From (2),
    Z = u^2 + v^2 = W^2
    Solve for u, v and W.
    Let (p,q) be positive integers, one odd, one even; p > q ; gcd(p,q) = 1 :

    Then we have two cases:
    Case 1: (p^2 - q^2) < 2pq (4)
    Case 2: (p^2 - q^2) > 2pq (5)

    =========================
    Case 1: (p^2 - q^2) < 2pq (4)
    =========================

    u = (2pq) (even)
    v = (p^2 - q^2) > 0 (odd)
    W = (p^2 + q^2) (odd)

    X = (u^2 - v^2) (odd) = 4p^2q^2 - (p^2 - q^2)^2 > 0
    Y = (2uv) (even) = 4pq(p^2 - q^2) > 0
    Z = (u^2 + v^2) (odd) = (p^2 + q^2)^2 = W^2

    (X + Y) is seen to be of the form 4*k_1 - (4*k_2 + 1) - 4k_3 = ( 4*k -
    1) and so cannot be a perfect square.
    So Case 1 is inadmissible.

    ==========================
    Case 2: (p^2 - q^2) > 2pq (5)
    ==========================

    u = (p^2 - q^2) > 0 (odd)
    v = (2pq) (even)
    W = (p^2 + q^2) (odd)

    X = (u^2 - v^2) (odd) = (p^2 - q^2)^2 - 4p^2q^2 > 0
    Y = (2uv) (even) = 4pq(p^2 - q^2) > 0
    Z = (u^2 + v^2) (odd) = (p^2 + q^2)^2 = W^2

    (X + Y) is seen to be of the form (4*k_2 + 1) - 4*k_1 - 4k_3 = ( 4*k
    +1) and so can be a perfect square.
    So Case 2 is admissible.

    (X+Y) = p^4 + q^4 - 6*p^2*q^2 + 4*p^3*q - 4*p*q^3 = W^2 (6)
    Or, equivalently,
    (p^2 - q^2 + 2*pq)^2 - 8*(pq)^2 = W^2 (7)

    The conditions are:
    p,q positive integers, one odd, one even; p > q ; (p^2 - q^2) > 2pq
    (8)

    Therefore, for fixed p,
    q^2 +2pq - p^2 < 0
    (q + p)^2 < 2*p^2
    q < (sqrt(2) - 1)*p
    q < 0.4142*p (9)

    Therefore, in principal, we could write a couple-of- lines program in
    Mathematica or Maple to obtain the smallest positive rational integer
    solution in a few milliseconds. We compute LHS of (7) for p = 1... N
    ; q = 1 ..... [0.4142*p] until it is a perfect square.

    The real task is to algebraically bootstrap our way up to the smallest
    rational integer solution from some starting poin, (which is what Chip has
    suggested.)

    regards, michael
     
    Michael, Dec 18, 2006
    #11
  12. Jim Dars

    Michael Guest

    Two questions come to mind:
    First, Jim Dars (and Fermat) asked for the smallest solution.
    In the reference cited, there is no proof that the solution found is the
    smallest one.
    Second, the approach gives one the the impression of being a 'hit or miss'
    affair and not a systematic one.
    --Michael
     
    Michael, Dec 18, 2006
    #12
  13. Jim Dars

    a1jrj Guest

    Excellent analysis - a short Mathcad program took two seconds to find
    the smallest solution of p :-
    p = 1469 q = 84
    with the same result as Fermat in the link. Which demonstrates it is
    the smallest solution.

    I think there is method in the cited link - basically a re-application
    of completing the square until the difference is a linear form which
    can be set to zero for a given rational number, but I agree it looks
    haphazard rather than systematical.

    JJ
     
    a1jrj, Dec 18, 2006
    #13
  14. Jim Dars

    Michael Guest

    I admit that there is method, which method perhaps falls under the umbrella
    of "Method of Diophantus".
    In addition to an appearance of being haphazard it does not seem to ensure
    that the smallest solution has been found.
    --michael
     
    Michael, Dec 18, 2006
    #14
  15. Jim Dars

    Proginoskes Guest

    Yes, but (1) Fermat found the answer systematically, and (2) Fermat
    didn't have Mathematica or Maple.

    --- Christopher Heckman
     
    Proginoskes, Dec 19, 2006
    #15
  16. Jim Dars

    Michael Guest

    Christopher, I do not *know* what *systematic* method Fermat used, which
    statement does not imply that he did not use a systematic one.
    --michael

     
    Michael, Dec 19, 2006
    #16
  17. Jim Dars

    sttscitrans Guest


    Yes, Gaussian integers give a quicker derivation

    x^2 +y^2 = z^4 =>
    (x +iy) = (p+iq)^4 =>
    x = p^4 -6p^2q^2 + q^4, y = 4ab(a^2-b^2) =>
    p^4 + q^4 - 6*p^2*q^2 + 4*p^3*q - 4*p*q^3 =w^2

    Assuming p,q, >0
    |x| + |y| = u^2 => sgn(x,y)=1 =>

    p^4 -6p^2q^2 + q^4 >0 and 4ab(a^2-b^2) >0
    or
    p^4 -6p^2q^2 + q^4 <0 and 4ab(a^2-b^2)<0

    p^4 -6p^2q^2 + q^4 =0 =>

    p/q = 1 +sqrt(2), -(1+sqrt(2), (1-sqrt(2), -(1-sqrt(2)

    => p/q > 1+sqrt(2) or sqrt(2)-1 < p/q <1

    If sgn(x,y) = -1, then you get solutions to the
    problem requiring the hypotenuse to be a square
    and the difference of the other two sides to be a square
    e.g 119^2 + 120^2 = 169^2

    Is there any reason to believe
    p^4 + q^4 - 6*p^2*q^2 + 4*p^3*q - 4*p*q^3 =w^2
    has an infinitude of integer solutions ?
    I have only found 1 solution for each case, although
    I have not searched very far.

    (p^2 - q^2 + 2*pq)^2 - 8*(pq)^2 = W^2 (7)

    This is a form of Pell's equation s^2 -2t^2 = 1
    If (p^2 - q^2 + 2*pq)^2 - 2*(2pq)^2 = 1

    Is there more than one right-angled triangle with sides
    n, n+1, m^2 ?
     
    sttscitrans, Dec 19, 2006
    #17
  18. Jim Dars

    Michael Guest

    Elegant!
    Glad to see you clarify the significance of Case 1.
    One of the joys of solving puzzles socially rather than onanistically is the
    diversity of perspectives one is exposed to.


    I believe that the current wisdom is that algebraic equations (with rational
    integer coefficients) of order greater than three, *in general*, tend to
    have none or very few rational integer solutions.



    This had occurred to me but I dismissed the thoughts, believing this to be a
    different and more difficult problem. I will look again for a possible
    connection.
    --michael
     
    Michael, Dec 19, 2006
    #18
  19. Jim Dars

    Jim Dars Guest

    Hi All,

    After a couple of days away I noticed in Michael's initial summation

    *********
    Given:
    (X,Y,Z,W) are positive integers; gcd(X,Y) = 1
    X^2 + Y^2 = Z^2 (1)
    Z= W^2 (2)
    X + Y = W^2 (3)
    **********************
    Shouldn't this read something like

    Given:
    (X,Y,Z,W) are positive integers; gcd(X,Y) = 1
    X^2 + Y^2 = Z^2 (1)
    Z= W^2 (2)
    and not
    X + Y = W^2 (3)
    but
    X + Y = E^2 (3) (assuming someone hasn't introduced E into the
    analysis.)??

    Best wishes, Jim
     
    Jim Dars, Dec 19, 2006
    #19
  20. Jim Dars

    Michael Guest


    Yes.
    Jim Dars you have a sharp eye!
    --michael.
     
    Michael, Dec 19, 2006
    #20
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