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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?

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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)
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    $\begingroup$ 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) $\endgroup$
    – justhalf
    Commented Sep 8, 2023 at 6:20
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    $\begingroup$ @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. $\endgroup$ Commented Sep 8, 2023 at 7:49
  • $\begingroup$ Well, fair point, I thought it's at least closer to a proof than simply stating it. But fine both ways :D $\endgroup$
    – justhalf
    Commented Sep 8, 2023 at 9:47

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