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Factors of triangular numbers forming arithmetic series

  1. Apr 28, 2006 #1
    For every pair (a,b) of factors that equal the trangular number m(m+1)/2 there are two distinct sets of pairs (c,d) that form a determinant equal to 2m+1 such that the products (a+cn)*(b+dn) = a triangular number for all n. Is this a previously known fact and how can it be prooved?
     
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  3. Apr 28, 2006 #2

    Hurkyl

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    T(n) = n²/2 + n/2

    T(kn + m) = (kn+m)²/2 + (kn+m)/2
    = (k²/2)n² + (km)n + (m²)/2 + (k/2)n + m/2
    = (k²/2)n² + (km+ k/2)n + (m² + m)/2
    = (k²/2)n² + (km + k/2)n + T(m)


    (a+cn)(b+dn) = (cd)n² + (ad + bc)n + ab



    Why look at the (kn+m)-th triangular number? Because I wanted the most general expression that was quadratic in n. (And later realized that one of my coefficients was your m)
     
  4. Apr 28, 2006 #3
    This is a helpful. I did determined the following formula for c and d that fits both the data and [tex]cd = k^2/2[/tex]
    Lets separate this into two parts m odd and m even
    For m even
    [tex]c_1 =(gcd(m+1,a))^{2}[/tex]
    [tex]d_1 =(gcd(m,b))^{2}*2[/tex]
    [tex]c_2 = (gcd(m,a))^{2}*2[/tex]
    [tex]d_2 =(gcd(m+1,b))^{2}[/tex]

    For m odd
    [tex]c_1 = (gcd(m+1,a))^{2}*2[/tex]
    [tex]d_1 =(gcd(m,b))^{2}[/tex]
    [tex]c_2 =(gcd(m,a))^{2}[/tex]
    [tex]d_2 =(gcd(m+1,b))^{2}*2[/tex]

    Either way [tex]cd = k^2/2[/tex] fits.

    Now to show that the data fits km + k/2 = ad + bc also with my equations. Granted that this is no proof but it could rule out my equations if it didn't. m = 36 m+1 = 37

    1. (1 + n)*(666+648n)
    gcd(666,36)=18; 2*1*18^2 = 648 -> k = 36
    n=1 -> 2*1314 = T(36+36)
    n=1 -> 3*1962 = T(72+36)
    36*36-36/2 = 1*648 + 666*1
    1314=1314

    2. (1 + 2n)*(666+1369n)
    gcd(666,37)= 37, 2*1*37^2 = 74^2/2 -> k = 74
    n=1 -> 3*2035 = T(74+36)
    n=2 -> 5*3404 = T(148+36)
    74*36+74/2 = 2701 = 1369 + 2*666

    3. (2+n)*(333+162n)

    , ...

    11. (18+n)*(37+2n)

    12. (18+648n)*(37+1369n)
     
    Last edited: Apr 29, 2006
  5. Apr 30, 2006 #4
    I check my data over and over and two principles remain.
    1. Although k can take any integer value in the first set of equations, there are only a finite number, i.e., [tex]\Upsilon(T(m))[/tex] of k values for which the diophantine equation set below has a solution in integers.
    [tex]ab = T(m)[/tex]
    [tex] cd = k^2/2[/tex]
    [tex]ad + bc = km + k/2[/tex]
    2. The solution for each k value is given by my equations for c and d as a function of a,b,m where [tex](a,b) \iff (m,k)[/tex] .
    [tex]\Upsilon(T(m))[/tex] equals the number of divisors of T(m).
    I am confident here that no counter example can be found.
     
  6. May 12, 2006 #5
    I have a proof for this statement. Anyone interested?
     
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