# The Triangular Cannonball Problem [closed]

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?

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)
• Do we need to add the short reasoning from that OEIS link? It's pretty short to include: For numbers to be triangular and tetrahedral, we look for solutions r*(r+1)*(r+2)/6 = t*(t+1)/2 = a(n). The corresponding r and t are r = A224421(n-1) and t = A102349(n). Writing m=r+1 and s=2t+1, this problem is equivalent to solving the Diophantine equation 3 + 4*(m^3 - m) = 3*s^2. The integer solutions for this equation are m = 0, 1, 2, 4, 9, 21, 35 and the corresponding values of s are 1, 1, 3, 9, 31, 111, 239. (End) Sep 8 at 6:20
• @justhalf I thought about it, but the proof isn't complete as it doesn't show why those are the only integer solutions for that Diophantine equation. Sep 8 at 7:49
• Well, fair point, I thought it's at least closer to a proof than simply stating it. But fine both ways :D Sep 8 at 9:47