# Triangular Number cannot be Cube

## Theorem

Let $T_n$ be the $n$th triangular number such that $n > 1$.

Then $T_n$ cannot be a cube.

## Proof

Suppose $T_n = x^3$ for some $x \in \Z$.

$\exists y \in \Z: 8 T_n + 1 = \paren {2 x}^3 + 1 = y^2$
$2 x = 2$, $y = 3$

giving the unique solution:

$T_n = 1^3 = 1$

$\blacksquare$