Determine the number of pairs of real numbers, $(x, y)$, with $0 \leq x \leq \frac{\pi}{8}$ and $0 \leq y \leq \frac{\pi}{8}$ and $\cos ^{6}(1000 x)-\sin ^{6}(1000 y)=1$.
Since $-1 \leq \cos \theta \leq 1$ for all angles $\theta$, then $0 \leq \cos ^{6} \theta \leq 1$ for all angles $\theta$.

Also, $\sin ^{6} \alpha \geq 0$ for all angles $\alpha$.

Therefore, to have $\cos ^{6} \theta-\sin ^{6} \alpha=1$ for some angles $\theta$ and $\alpha$, then we must have $\cos ^{6} \theta=1$ and $\sin ^{6} \alpha=0$.

From the given equation, this means that $\cos ^{6}(1000 x)=1$ and $\sin ^{6}(1000 y)=0$.

Therefore, $\cos (1000 x)= \pm 1$ and $\sin (1000 y)=0$.

Since $0 \leq x \leq \frac{\pi}{8}$, then $0 \leq 1000 x \leq 125 \pi$.

Now $\cos \theta= \pm 1$ exactly when $\theta$ is a multiple of $\pi$.

Therefore, the possible values of $1000 x$ are $0, \pi, 2 \pi, \ldots, 124 \pi, 125 \pi$.

There are 126 such values.

Since $0 \leq y \leq \frac{\pi}{8}$, then $0 \leq 1000 y \leq 125 \pi$.

Now $\sin \alpha=0$ exactly when $\alpha$ is a multiple of $\pi$.

Therefore, the possible values of $1000 y$ are $0, \pi, 2 \pi, \ldots, 124 \pi, 125 \pi$.

There are 126 such values.

Since there are 126 values of $x$ and 126 independent values of $y$, then there are $126^{2}=15876$ pairs $(x, y)$ that satisfy the original equation and restrictions.