Reference: the first $m$ triangular numbers are a complete residue system modulo $m$ iff $m$ is a power of $2$

Question

Consider, for any $n\in\mathbb{Z}$, the $n$th-triangular number $T_{n}=\frac{n(n+1)}{2}$. I just noticed that, for $m\in\mathbb{N}\setminus\{0,1\}$, $$T_{0}, T_{1}, ... , T_{m-1}$$ is a complete residue system modulo $m$ (meaning, as it always does, that for every $i\in\{1, 2,... , m\}$ there exists $j$ on the same set such that $T_{j}\equiv i$ modulo $m$) if, and only if, $m$ is a power of $2$.

Just for illustration:

• for $m=2$, $T_{0}=0\equiv 0$ and $T_{1}=1\equiv 1$;
• for $m=3$, $T_{0}\equiv 0$, $T_{1}\equiv 1$, and $T_{2}=3\equiv 0$;
• for $m=4$, $T_{0}\equiv 0$, $T_{1}\equiv 1$, $T_{2}\equiv 3$, $T_{3}=6\equiv 2$;
• for $m=5$, $T_{0}\equiv 0$, $T_{1}\equiv 1$, $T_{2}\equiv 3$, $T_{3}\equiv 1$ and $T_{4}=10\equiv 0$;
• and so on...

I am not a number theorist, so the proof I was able to come up with for this fact is not straightforward and very long... What would be a standard, preferably short proof of this? And can it be easily generalized to pentagonal numbers, hexagonal numbers and so on?

Answer 1

A short proof.

First notice that for any $i, j$, we have $T(i) \equiv T(j) \pmod m \iff 2m\mid (i - j)(i + j + 1)$.

We can write $m = 2^r t$ with $t$ odd.

We take $i = \frac{t + 2^{r + 1} - 1}2$ and $j = \frac{|t - 2^{r + 1}| - 1}2$ and verify that $(i - j)(i + j + 1) = 2^{r + 1}\cdot t = 2m$.

If $t > 1$, then both $i, j$ are in the range $0, \dots, m - 1$ and are different. Therefore $T(0), \dots, T(m - 1)$ cannot be a complete residue system mod $m$.

Now suppose $t = 1$ and hence $m = 2^r$. It suffices to show that, for any $0 \leq j < i < m$, we have $2^{r + 1} \nmid (i - j)(i + j + 1)$.

But the two factors $i - j$ and $i + j + 1$ have different parity, hence one of them is odd; and the other lies in the interval $(0, 2^{r + 1})$. This finishes our proof.