A Game on Triangular and Square Numbers

Question

A game is played by $n$ people $A_1, A_2,...,A_n$ with the following rules:

1. The $n$ people take turn to call out a positive number, in ascending order,
$\quad$i.e. $A_1$ says 1, $A_2$ says 2, $...$ , $A_n$ says $n$, $A_1$ says $(n+1)$, $A_2$ says $(n+2)$ and so on.

2. When a person calls out a triangular number or a square number (or both at the same time), he needs to clap his hands.

If $n=2$, both of the two players have the chance to clap. However, if $n=3$, one of the players never needs to clap. I have proved this case using modular arithmetic.

My question is, what are the values of $n$ such that there is at least one player never needs to clap? Is 3 the only answer?

Answer 1

We have $n$ players.
The $i$-th player says the numbers of the form $i+n\cdot m$ where $m=1,2,3,...$
When $n=3$ the second player will not clap his hands since we cannot have
$x^2\equiv 2 \pmod 3$ or $\frac{y(y+1)}{2}\equiv 2 \pmod 3$ as you proved.
But the second player will have the same problem when $n$ is any multiple of $3$ since the same congruences must also hold.
Therefore, $n=3$ is not the only case.

Answer 2

If $n$ is any odd $p$ there exists at least a single slot, $i$, that never gets either a square nor a triangular number. to see this:

Note that $i$ gets a square if and only if $i$ is a quadratic residue mod($p$). $i$ gets a triangular number if and only if there is some residue $k$ with $k^2+k\equiv 2i$ mod($p$). That quadratic congruence will have a solution if and only if $1+8i$ is a quadratic residue (by the quadratic formula). Thus we are down to showing that there is at least a single residue $i$ such that neither $i$ nor $1+8i$ is a quadratic residue mod ($p$).

To see this, suppose it were not the case. Then for every non-square, $i$ it must be that $1+8i$ is a square. Counting $i=0$ there is exactly one more square than non-square. But if $i=0$ then $1+8i=1$ is a square and if $i=1$ then $1+8i=9$ is a square, hence there aren't enough squares to pair with the non-residues so, by the pigeonhole principle, we are done.