Dots in Triangles and Pyramids

1. The triangular numbers count dots in a triangular array, $1, 3, 6, 10, \dots$, the $n$th being $1 + 2 + \dots + n$. If a triangle is built from $2701$ dots, how many more dots are needed for the next larger triangle?

2. The numbers $1$ and $36$ are each at once a triangular number and a perfect square. What are the next four numbers that are both?

3. Stacking triangles gives the tetrahedral numbers $1, 4, 10, 20, \dots$, the $n$th being the sum of the first $n$ triangular numbers. Show that two consecutive tetrahedral numbers add to a square pyramidal number (the count of cannonballs in a square-based pyramid, $1^2 + 2^2 + \dots + n^2$). Then find every square pyramidal number, other than $1$, that is also a perfect square.

Solution

Part 1. The $n$th triangular number is $n(n+1)/2$. Setting this to $2701$ gives $n(n+1) = 5402$, so $n = 73$ (since $73 \times 74 = 5402$). The next triangle simply adds one more row, of $74$ dots. So $74$ more are needed.

Part 2. The next four square-triangular numbers after $1$ and $36$ are $$\begin{gathered} 1225 = 35^2, \quad 41616 = 204^2,\\ 1413721 = 1189^2, \quad 48024900 = 6930^2, \end{gathered}$$ and each is also triangular ($T_{49}$, $T_{288}$, $T_{1681}$, $T_{9800}$). They obey the tidy rule $a_{n+1} = 34 a_n - a_{n-1} + 2$: for instance $34 \times 36 - 1 + 2 = 1225$ and $34 \times 1225 - 36 + 2 = 41616$.

Part 3. Writing the tetrahedral number as $T\!e_n = n(n+1)(n+2)/6$, $$\begin{aligned} T\!e_n + T\!e_{n-1} &= \frac{n(n+1)(n+2)}{6} + \frac{(n-1)n(n+1)}{6}\\ &= \frac{n(n+1)\bigl[(n+2)+(n-1)\bigr]}{6} = \frac{n(n+1)(2n+1)}{6}, \end{aligned}$$ which is exactly $1^2 + 2^2 + \dots + n^2$, the square pyramidal number. As for which of these are perfect squares: apart from the trivial $1$, the only one is $$4900 = 70^2 = 1^2 + 2^2 + \dots + 24^2.$$ This is the famous cannonball problem, that a square layer of $70 \times 70$ holds exactly as many as a square pyramid $24$ high. That $4900$ is the sole non-trivial solution was proved by G. N. Watson in $1918$, so we take it as settled rather than reprove it here.