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Homework Help: Sum of Two Triangular Numbers

  1. Mar 22, 2006 #1

    AKG

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    Which number can be expressed as the sum of two triangular numbers? I don't even know how to start with this one. Here is some data:

    If you don't count 0 as a triangular number, then the following can:

    2
    4
    6
    7
    9
    11
    12
    13
    16
    18
    20
    21
    22
    24
    25
    27
    29
    30
    31
    34
    36
    37

    And these can't:

    3
    5
    8
    10
    14
    15
    17
    19
    23
    26
    28
    32
    33
    35

    If you do count 0 as one, then the following can:

    2
    3
    4
    6
    7
    9
    10
    11
    12
    13
    15
    16
    18
    20
    21
    22
    24
    25
    27
    28
    29
    30
    31
    34
    36
    37

    And these can't:

    5
    8
    14
    17
    19
    23
    26
    32
    33
    35

    Unlike the sum of two squares problem, you don't get anything so nice like if a and b are the sum of two triangles, then so is ab. Also, with sums of two squares, there's the fact if p is an odd prime, then p is a sum of two squares iff p = 1 (mod 4). Again, nothing as nice appears to be true for triangle numbers (even if you replace mod 4 with mod 3 or other small primes, or so it seems).
     
  2. jcsd
  3. Mar 22, 2006 #2

    Hurkyl

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    One thing that strikes me about this list is that it has:
    5, 8
    14, 17 (= 5 + 9, 8 + 9)
    23, 26 (= 14 + 9, 17 + 9)
    32, 35 (= 23 + 9, 26 + 9)
    With 19 and 33 the only ones remaining.

    Maybe it's just a spurious pattern that arises because we're only looking at very small numbers, though.
     
  4. Mar 22, 2006 #3

    Hurkyl

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    Another thing that may or may not be useful: observe that the sum of the m-th and n-th triangular numbers is:

    [tex]\frac{m(m+1)}{2} + \frac{n(n+1)}{2}
    = \frac{1}{2} \left( \left(m + \frac{1}{2}\right)^2 + \left(n + \frac{1}{2}\right)^2 - \frac{1}{2} \right)[/tex]

    Maybe you could apply some of the reasoning for the sum of two squares to this case, through an affine transformation.
     
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