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?

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)

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

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.