The Triangular Cannonball Problem [closed]

Question

How many ways are there to stack an equilateral triangle of cannonballs into a tetrahedron of cannonballs? In other words, how many positive integers are both triangular and tetrahedral?

Answer 1

The $n$th triangular number is $\frac{n(n+1)}{2}=\binom{n+1}{2}$.

The $n$th tetrahedral number is $\frac{n(n+1)(n+2)}{3}=\binom{n+2}{3}$.

Only the following numbers are both tetrahedral and triangular, i.e. there are only four possible pairs $(n,m)$ of positive integers with $\binom{n+1}{2}=\binom{m+1}{3}$:

• $1=\binom{1+1}{2}=\binom{1+2}{3}$ (this is the $1$st triangular number and $1$st tetrahedral number, a trivial case)
• $10=\binom{4+1}{2}=\binom{3+2}{3}$ (this is the $4$th triangular number and $3$rd tetrahedral number)
• $120=\binom{15+1}{2}=\binom{8+2}{3}$ (this is the $15$th triangular number and $8$th tetrahedral number)
• $1540=\binom{55+1}{2}=\binom{20+2}{3}$ (this is the $55$th triangular number and $20$th tetrahedral number)
• $7140=\binom{119+1}{2}=\binom{34+2}{3}$ (this is the $119$th triangular number and $34$th tetrahedral number)