[Geometry]
Problem: Suppose that $\triangle A B C$ is right-angled at $B$ and has $A B=n(n+1)$ and $A C=(n+1)(n+4)$, where $n$ is a positive integer. Determine the number of positive integers $n<100000$ for which the length of side $B C$ is also an integer.

Solution: Since $\triangle A B C$ is right-angled at $B$, then

$$
\begin{aligned}
B C^{2} & =A C^{2}-A B^{2} \\
& =((n+1)(n+4))^{2}-(n(n+1))^{2} \\
& =(n+1)^{2}(n+4)^{2}-n^{2}(n+1)^{2} \\
& =(n+1)^{2}\left((n+4)^{2}-n^{2}\right) \\
& =(n+1)^{2}\left(n^{2}+8 n+16-n^{2}\right) \\
& =(n+1)^{2}(8 n+16) \\
& =4(n+1)^{2}(2 n+4)
\end{aligned}
$$

The length of $B C$ is an integer exactly when $4(n+1)^{2}(2 n+4)$ is a perfect square.

Since $4(n+1)^{2}$ is a perfect square, then $B C$ is an integer exactly when $2 n+4$ is a perfect square.

We note that $2 n+4 \geq 6$ (since $n \geq 1)$ and that $2 n+4$ is even.

Since $n<100000$, then $6 \leq 2 n+4<200004$, and so we need to count the number of even perfect squares between 6 and 200004 .

The smallest even perfect square in this range is $4^{2}=16$.

Since $\sqrt{200004} \approx 447.2$, the largest even perfect square in this range is $446^{2}$.

Therefore, the number of even perfect squares in this range is $\frac{446}{2}-1=222$.

Thus, there are 222 positive integers $n$ for which the length of $B C$ is an integer.

Final answer: 222