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

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle r \left({r+1}\right)\)	\(=\)	\(\displaystyle \left({3 k + 1}\right) \left({3 k + 2}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 9 k^2 + 9 k + 2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 9 k \left({k + 1}\right) + 2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

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

Thus $T_r \equiv 1 \pmod 9$.

$\blacksquare$