Triangular Number Modulo 3 and 9

Theorem
Let $n$ be a triangular number.

Then one of the following two conditions applies:
 * $n \equiv 0 \pmod 3$
 * $n \equiv 1 \pmod 9$

Proof
Let $n = T_r$.

Then $n = \dfrac {r \left({r+1}\right)} 2$ from Closed Form for Triangular Numbers.

It suffices to investigate the nature of $r \left({r+1}\right)$ modulo $3$ from Euclid's Lemma.

There are three cases to consider:
 * 1) $r \equiv 0 \pmod 3$
 * 2) $r \equiv 1 \pmod 3$
 * 3) $r \equiv 2 \pmod 3$


 * Let $r \equiv 0 \pmod 3$.

Then $r \left({r+1}\right) \equiv 0 \pmod 3$ and so $T_r \equiv 0 \pmod 3$.


 * Let $r \equiv 2 \pmod 3$.

Then $r+1 \equiv 3 \equiv 0 \pmod 3$.

So $r \left({r+1}\right) \equiv 0 \pmod 3$ and so $T_r \equiv 0 \pmod 3$.


 * Let $r \equiv 1 \pmod 3$.

Then $\exists k \in \Z: r = 3 k + 1$.

So $r \left({r+1}\right) = \left({3 k + 1}\right) \left({3 k + 2}\right)$

So $T_r = 9 \dfrac {k \left({k + 1}\right)} 2 + 1$.

Thus $T_r \equiv 1 \pmod 9$.