2008 AIME II Problem 15

Condition $\mathrm{(i)}$ says $n^2 = 3m^2 + 3m + 1,$ which rearranges to $(2n - 1)(2n + 1) = 3(2m + 1)^2$

The two factors are coprime, and working modulo $3$ shows $2n - 1$ must be a perfect square; then $2n + 79 = d^2$ yields a difference of squares equal to $80$

Condition $\mathrm{(i)}$ says $n^2 = (m + 1)^3 - m^3$ $= 3m^2 + 3m + 1$ for some integer $m.$ Multiplying by $4$ and rearranging, $4n^2 - 1 = 12m^2 + 12m + 3,$ i.e. $$(2n - 1)(2n + 1) = 3(2m + 1)^2.$$ The factors on the left are consecutive odd numbers, hence coprime, so one of them is a perfect square and the other is $3$ times a square. If $2n - 1 = 3k^2,$ then $2n + 1 \equiv 2 \pmod 3$ would be a perfect square, which is impossible. Hence $2n - 1 = k^2$ with $k$ odd.

Writing $k = 2a + 1$ gives $n = 2a^2 + 2a + 1.$ Condition $\mathrm{(ii)}$ says $2n + 79 = 4a^2 + 4a + 81 = d^2,$ so $$\begin{aligned} &(d - 2a - 1) \\ &\quad {}\cdot (d + 2a + 1) \\ &= d^2 - (2a + 1)^2 \\ &= 80. \end{aligned}$$ The two factors have the same parity, so both are even: the pairs $(2, 40),$ $(4, 20),$ $(8, 10)$ give $2a + 1 = 19,$ $8,$ $1,$ of which the odd values yield $a = 9$ (so $n = 181$) and $a = 0$ (so $n = 1$).

For $n = 181:$ indeed $181^2 = 32761 = 105^3 - 104^3$ (here $2n + 1 = 363 = 3 \cdot 11^2,$ as required), and $2n + 79 = 441 = 21^2.$ So the largest such $n$ is $181.$