Question

Show that no triangular number can be of the form $3n-1$.

Answer 1

Let $\frac{x(x+1)}{2}$ be the triangular number, then this is a multiple of $3$ if either $x$ or $x+1$ is divisible by $3$. If neither $x$ nor $x+1$ is divisible by $3$.

$x$ must be of the form $3n+1$;

In this case $\frac{1}{2}x(x+1)=\frac{9}{2}n(n+1)+1$ and is therefore of the form $3r+1$.

Thus the form $3n-1$ is not possible.