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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

1755.0. "College Mathematics Journal #501" by RUSURE::EDP (Always mount a scratch monkey.) Wed May 26 1993 19:36

    Proposed by K. R. S. Sastry, Addis Ababa, Ethiopa.
    
    Let ABC be an integer-sided right triangle, with C = 90 degrees.  The
    g.c.d. of the lengths of its sides a, b, c is one.  Prove or disprove
    that its semi-perimeter is a triangular number if and only if either a
    and c or b and c are consecutive integers.
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1755.1Conjecture in .0 is TRUETROOA::RITCHEFrom the desk of Allen Ritche...Sat Jun 05 1993 17:59104
    
    I believe all right-angled triangles with integer sides can be found
    using:
    
      a = k*2xy            b = k (x^2 - y^2)         c = k (x^2 + y^2)
    
    where k, x, and y are positive integers, and x > y.
    And GCD (x,y)=1
    
    If GCD(a,b,c)=1 then k=1 so the semi-perimeter
    
    
         S = 1/2 (2xy + x^2 - y^2 + x^2 + y^2)
    
           = x (x + y)
    
    
    Now if b and c are to be consecutive integers...
    
      c - b = 1
    
      x^2 + y^2 - x^2 + y^2  = 1
    
      2y^2 = 1  which has no integral solution.
    
    This should have been more obvious to me since the longest
    side and shortest side cannot possibly be consecutive!
    
    
    But if a and c (the longest two sides) are consecutive integers...
    
      c - a = 1
    
      x^2 + y^2 - 2xy = 1
    
      (x-y)^2 = 1
    
      x - y = +1  which means y must be (x - 1) with x>=2
    
    (x and y are consecutive! not to be confused with a,b,c)
    
    
    Finally, the semi-perimeter
    
       S = x (x + y)
         = x (x + x - 1)
         = x (2x - 1)
    
    
    A triangle number is T(n) = SUM [i], i=1,n  = n(n+1)/2
    
      Let n = 2x - 1   x>=2
    
      Then T(n) = (2x-1)(2x)/2 = x(2x-1) = S
    
    
    So the semi-perimeter starts with T(3) and every other one thereafter.
    ("odd" triangle numbers)
    
    E.g.
    
    x=2 n=3 yields triangle (4,3,5)     S=6
    x=3 n=5 yields triangle (12,5,13)   S=15
    x=4 n=7 yields triangle (24,7,25)   S=28
    x=5 n=9 yields triangle (50,9,49)   S=54
    etc.
    
    It appears that in a triangle defined by .0, the semi-perimeter is
    a triangular number iff the two longer sides are consecutive integers.
    Symmetry explains the "either... or" clause in .0
    
    I've only proved half of the iff, however.  i.e. if consec then
    triangle.
    
    
    To prove the converse: if S is triangle number then consec is not so
    obvious...
    
    For odd triangle numbers, T(odd), it is trivial to show that
    x - y must be 1 and therefore c - a = 1.  Note that x and y are
    consecutive if and only if c and a are consecutive.
    
    For T(even), say T(2x) without loss of generality, I try to
    solve T(2x)=x(x+y)
    
      2x(2x+1)/2 = x(x+y)
    
      2x+1 = x + y
    
      x = y - 1
    
    This contradicts x>y as stated at the beginning.  So this implies that
    that the S of a right triangle, if a triangle number, must
    be an odd triangle number.
    
    Therefore, "if S is triangle then consec" is true.
    
    And therefore it is *PROVEN* that in a right-angled triangle, the
    semi-perimeter is a triangular number if and only if the two longer
    sides are consecutive integers.
    
      
    Allen