Cubed Triangles

Suppose a number $N$ is both cubic and triangular, i.e., $N = m^{3}$ for some positive integer $m$ and $N = \dfrac{n(n + 1)}{2}$ for some positive integer $n$ . Then

$m^{3} = \dfrac{n(n + 1)}{2} \Longrightarrow 2m^{3} = n^{2} + n \Longrightarrow 8m^{3} = 4n^{2} + 4n + 1 - 1 = (2n + 1)^{2} - 1 \Longrightarrow (2n + 1)^{2} - (2m)^{3} = 1$ .

But by Catalan's Conjecture/Theorem the only consecutive positive ( $\ge 2$ ) powers are $2^{3}$ and $3^{2}$ , implying that $2n + 1 = 3 \Longrightarrow n = 1$ , and $2m = 2 \Longrightarrow m = 1$ . In both cases we end up with $N = 1$ , verifying that this is the only cubic, triangular number.