What is the modulus for any Tetrahedral number n by any Triangular number m? Given that the series of Tetrahedral numbers is defined by 
a(n) = C(n+2,3) = n*(n+1)*(n+2)/6
https://oeis.org/search?q=1%2C4%2C10%2C20%2C35&sort=&language=&go=Search
and the series of Triangular numbers is defined by 
a(m) = binomial(m+1,2) = m(m+1)/2
https://oeis.org/search?q=1%2C3%2C6%2C10%2C15%2C21&sort=&language=&go=Search
and given the unique geometries of these numbers...
Tetrahedral: "number of balls in a triangular pyramid in which each edge contains n balls". 
Triangular:  "number of balls in a triangle with base m".
...you would think that there would be an elegant relationship between these two types of numbers that would enable you to easily calculate the quotient and remainder when a Tetrahedral number n is divided by a Triangular number m (n and m not equal).  I am looking for an algebraic pattern, not a simple methodology.
There is at least one case where the modulus for a given n and m is 0: Tetrahedral(8) mod(Triangular(15)) = 0 ; both values are 120.
As a starting point, for any given Triangular number m, what is the closet Tetrahedral number n and what is the modulus?  Is there an algebraic pattern to these moduli?  As m increases by 1 what is the closest n and what is the modulus?  
Or another way to look at the problem:  As n increases by 1 what is the closest m and what is the modulus?
Note:  I am an engineer with an interest in Number Theory.  I am thankful for the insights of those with greater experience.
 A: "... you would think that ... " unfortunately does not work very well in mathematics. Apparent simple problems are sometimes difficult or impossible to solve, and complex problems can sometimes be rephrased and solved by some clever observations.
You might for instance expect that there are many integers that are both a triangular as well as a tetrahedral number. The first few are 1,10,120,1540,7140, which are the 1st,3rd,8th,20st,34th tetrahedral numbers and the 1st,4th,15th,55th,119th triangular numbers, but there are no other solutions for $n<10^6$.
Their sequence is listed https://oeis.org/A224421 , and does not contain any other members. They appear to be the only solutions, but unfortunately the reference that probably contains the proof of that is in Russian.
It is of course easy to get the solution for 
$$
\frac{m(m+1)}{6} = \frac{n(n+1)(n+2)}{12}
$$ 
and the index of the nearest triangular number for given $n$ is either rounded up or down of
$$
\frac{-3 + \sqrt{12 n(n+1)(n+2) +9}}{6}
$$
You can also do this for $n$ as function of $m$, except that solving a third order equation gives a more ugly expression.
So it is easy to calculate the nearest pairs and differences, but I don't think that there is a particular appealing or nice pattern to obtain for it unless you would like an asymptotic expansion in $\sqrt{n}$ or $^3\sqrt{m}$.
