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 $x \in \Z$.

Then by Odd Square is Eight Triangles Plus One:

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

By Cube which is One Less than a Square:

$2 x = 2$, $y = 3$

giving the unique solution:

$T_n = 1^3 = 1$

$\blacksquare$

Sources

• 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $15$
• 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $15$