Context:
A triangular number is a number of the form $n(n+1)/2$ where $n\in \mathbb N$.
Question:
After application of two theorems in my textbook (the first one being that "a number $n$ is triangular if and only if $\exists N\in \mathbb N,\text{ }8n+1 = N^2$" the second one being "if a number $k$ is odd then $\exists p \in \mathbb N, \text{ } k^2 = 8p + 1$") the following result came out:
To every odd number $k$, we can assign a triangular number $n$, namely: $\frac{k^2-1}{8}$.
My question is two-fold:
- does the converse hold? That is, can we assign to every triangular number an odd number?
- if the answer of question $1$ is yes, is there by any chance, a way to assign to a every even number a special number in direct analogy with the way it happens for odd numbers?
Thanks
Note that $$\frac {k^2-1}8=\frac 12 \cdot \frac {k-1}2 \cdot \frac {k+1}2$$ and setting $k=2n+1$ we obtain $$\frac {k^2-1}8=\frac {n(n+1)}2$$
So if we have a triangular number we can assign the odd number $k=2n+1$ and the arithmetic all works.
It is artificial to do anything with even numbers in the same way, because essentially you would be looking for even numbers to map to somewhere in the gaps between triangular numbers. You can, of course simply plug $k=2n$ into the formula you have, but that will not give you an integer. If you have to change the formula to get what you want, that is an indication that the generalisation may not be fruitful.