, 2013

Solution:

Answer: 761474

As an approximation, we assume $a, b, c, d$ are ordered to begin with (so we have to divide by $24$ later) and add to $168$ with a unique choice of signs; then, it suffices to count $e + f + g + h = 168$ with each $e, f, g, h$ in $[-168, 168]$ and then divide by $24$ (we drop the condition that none of them can be zero because it shouldn't affect the answer that much).

One way to do this is generating functions. We want the coefficient of $t^{168}$ in the generating function $\left(t^{-168} + t^{-167} + \ldots + t^{167} + t^{168}\right)^{4} = \left(t^{169} - t^{-168}\right)^{4} / (t-1)^{4}$.

Clearing the negative powers, it suffices to find the coefficient of $t^{840}$ in
$$\frac{\left(t^{337} - 1\right)^{4}}{(t-1)^{4}} = \left(1 - 4 t^{337} + 6 t^{674} - \ldots\right) \frac{1}{(t-1)^{4}}.$$
To do this we expand the bottom as a power series in $t$:
$$\frac{1}{(t-1)^{4}} = \sum_{n \geq 0} \binom{n+3}{3} t^{n}$$
It remains to calculate $\binom{840+3}{3} - 4 \cdot \binom{840-337+3}{3} + 6 \cdot \binom{840-674+3}{3}$. This is almost exactly equal to $\frac{1}{6}\left(843^{3} - 4 \cdot 506^{3} + 6 \cdot 169^{3}\right) \approx 1.83 \times 10^{7}$.

Dividing by $24$, we arrive at an estimation $762500$. Even if we use a bad approximation $\frac{1}{6 \cdot 24}\left(850^{3} - 4 \cdot 500^{3} + 6 \cdot 150^{3}\right)$ we get approximately $933000$, which is fairly close to the answer.