Piggy Bank Diet

In a square pyramid of coins, the $n$ th layer (counting from the top) would contain $n^2$ coins. In a tetrahedral pyramid, the same layer would contain $T_n$ coins, where $T_n$ is the $n$ th triangular number, whose formula is $\frac{n(n + 1)}{2}$ .

Let $SP_m$ and $TP_n$ be the number of coins in a square pyramid and a tetrahedral pyramid with $m$ and $n$ layers, respectively. Then

$\begin{array}{rl} SP_m &= \ \ 1 + 4 + 9 + \ldots + m^2 \\ \\ &= \ \ \displaystyle \sum_{k = 1}^m k^2 \\ \\ &= \ \ \dfrac{m(m + 1)(2m + 1)}{6} \\ \\ \\ TP_n &= \ \ 1 + 3 + 6 + 10 + \ldots + T_n \\ \\ &= \ \ \displaystyle \sum_{k = 1}^n \dfrac{k(k + 1)}{2} \\ \\ &= \ \ \dfrac{1}{2} \left( \displaystyle \sum_{k = 1}^n k^2 + \displaystyle \sum_{k = 1}^n k \right) \\ \\ &= \ \ \dfrac{1}{2} \left( \dfrac{n(n + 1)(2n + 1)}{6} + \dfrac{n(n + 1)}{2} \right) \\ \\ &= \ \ \dfrac{n(n + 1)}{2} \left( \dfrac{2n + 1}{6} + \dfrac{1}{2} \right) \\ \\ &= \ \ \dfrac{n(n + 1)(n + 2)}{6} \end{array}$

So we are seeking two sets of $(m, n)$ such that $SP_m + TP_n$ are perfect cubes no greater than 5000.

This seems difficult (impossible?) to solve analytically, so we resort to a program and find three sets of values for $(m, n)$ : $(1, 19), (2, 8)$ and $(16, 24)$

We discard the first as $m = 1$ would be a single coin and not qualify as a pyramid.

This leaves $SP_2 + TP_8 = 125 = 5^3$ and $SP_{16} + TP_{24} = 4096 = 16^3$ , so the amount spent was $4096 - 125 = \boxed{3971}$

4096-125=3971