My question is motivated by the following fact (for a couple of proofs see my answer at MSE 1926967; note that this fact also answers the $\ell = 1$ case of MO 265513):

Fact. The set of triangular numbers modulo $n$ yields $\mathbb{Z}/n\mathbb{Z}$ iff $n$ is a power of $2$.

My specific question is whether there is a straightforward way to describe this set given, for example, the prime factorization of $n$. But I also have the broader question:

What else can be said about the set of triangular numbers modulo $n$ as a function of $n$?

For example, when $n$ is odd, we might expect the set to have cardinality around $(n+1)/2$. To illustrate this by way of example, consider $n = 11$, and the first $11$ counting numbers each reduced modulo $11$:

$$1, 2, 3, 4, 5, -5, -4, -3, -2, -1, 0$$

However many distinct elements we have after the first $(n+1)/2$ triangular numbers will be the final total, since (as suggested by the negative notation above) the elements reverse course thereafter, and wend their way back to $0$ at which point they reset as they have formed a cycle.

But I am not only interested in an estimate of the cardinality based on $n$'s parity; I am hoping for the whole shebang! In that spirit:

Q. How can we describe the set of triangular numbers modulo $n$ given $n$'s prime factorization?


Fix $n=p_1^{\alpha_1}\ldots p_k^{\alpha_k}$. We want to know for which $b$ the equation $$ a(a+1)/2=b $$ has a solution modulo $n$. By the Chinese remainder theorem, this happens iff it has a solution mod $p_i^{\alpha_i}$ for each $i$. Consider one of these prime powers. If $p_i=2$ then there is always a solution, as mentioned in your question. Suppose $p_i\neq2$. The equation is equivalent to $$ (2a+1)^2=8b+1. $$ Let $8b+1=x_ip_i^{\beta_i}$ where $p_i\not\mid x_i$. If $\beta_i\geq\alpha_i$ we can take $2a+1=p^{\alpha_i}$. Otherwise $\beta_i$ must be even, and in this case by Hensel's lemma, there is a solution iff $x_i$ is a quadratic residue mod $p_i$. Therefore the residues of triangular numbers mod $n$ are exactly $$ \left\{b\,\left|\,\left(\frac{x_i}{p_i}\right)=1\text{ and }\beta_i\text{ is even for each }i\text{ with }\beta_i<\alpha_i\text{ and }p_i\neq2\right.\right\} $$ where $\left(\frac{x}{p}\right)$ is the Legendre symbol. I don't think there is a closed form expression for the Legendre symbol, but quadratic reciprocity gives an algorithm for calculating it.

  • $\begingroup$ Awesome. Three questions: (a) What happens in the case of $n$ even but not a power of $2$? (b) Does the re-framing of these residues using the Legendre symbol speed anything up computationally in generating this set (as compared to, e.g., directly computing them as sums modulo $n$)? (c) For a sample consideration such as the one initially posted -- expected cardinality of the set for some given $n$ and its prime factorization -- is there a way to carry out this estimate using the methods you outline here? $\endgroup$ – Benjamin Dickman Sep 27 '16 at 0:33
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    $\begingroup$ (a) The last equation is still valid when $p_1=2$, there is just no condition imposed by $p_1$. That is, $b$ is a triangular number mod $n$ iff it is triangular mod $n'$ where $n'$ is the odd part of $n$. (b) My guess is computing sums modulo $n$ is faster if you want all the residues, while the above would be better for testing a particular number $b$. (c) It should be possible to calculate using these methods (there are $(p+1)/2$ quadratic residues mod $p$ including $0$, when $p$ is an odd prime). $\endgroup$ – stewbasic Sep 27 '16 at 23:58
  • $\begingroup$ Actually, the equation is equivalent to $(2a+1)^2=8b+1 \pmod{p_i^{\alpha_i}}$. If $8b+1$ is a quadratic residue, then we can always solve for $a$, since $(8, p_i^{\alpha_i})=1$ $\endgroup$ – steven gregory Aug 17 '17 at 14:52
  • $\begingroup$ @stevengregory Yes, I think this is implicit in my answer; the condition "$\beta_i\geq\alpha_i$ or ($\left(\frac{x_i}{p_i}\right)=1$ and $\beta_i$ is even)" is equivalent to $8b+1$ being a quadratic residue mod $p_i^{\alpha_i}$. $\endgroup$ – stewbasic Aug 17 '17 at 22:10

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