Mysterious Integer

Let $p=2n+1$ . Then $\dfrac{n}{3}(4n^2 +6n +3) = \dfrac{2n}{6}(4n^2 +4n +1+(2n +1) +1) = \dfrac{p-1}{6}(p^2+p +1) = \dfrac{p^3-1}{6}$ .

Then for a triangular number equality, let $\dfrac{p^3-1}{6} = \dfrac{m(m+1)}{2}$ for some $m$ .

Then $p^3 - 1 = 3m^2 + 3m$ .

$p^3 + m^3 = m^3 + 3m^2 + 3m + 1 = (m+1)^3$ .

However, according to Fermat's last theorem , no integer solutions for the equation in a form of $x^n + y^n = z^n$ .

Therefore, there are no integer solutions $n$ that can satisfy all constraints.

A triangular number has the form $\frac{n(n+1)}{2}$ , hence we require:

$\frac{n(n+1)}{2} = \frac{4n^3 + 6n^2 + 3n}{3} \Rightarrow 3n^2 + 3n = 8n^3 + 12n^2 + 6n \Rightarrow 0 = 8n^3 + 9n^2 + 3n$

or $0 = n(8n^2 + 9n + 3)$ (i)

The quadratic term is irreducible and has roots $n = \frac{-9 \pm \sqrt{81 - 4(8)(3)}}{16} = \frac{-9 \pm \sqrt{15}i}{16}$ , as well as $n = 0$ is the third remaining root in (i) . There are NO values $n \in \mathbb{N}$ that satisfy the triangular number condition.