For each positive integer $n$, define the point $P_{n}$ to have coordinates $\left((n-1)^{2}, n(n-1)\right)$ and the point $Q_{n}$ to have coordinates $\left((n-1)^{2}, 0\right)$. For how many integers $n$ with $2 \leq n \leq 99$ is the area of trapezoid $Q_{n} P_{n} P_{n+1} Q_{n+1}$ a perfect square?
We first note that the coordinates of $P_{n+1}$ are $\left(((n+1)-1)^{2},(n+1)((n+1)-1)\right)=\left(n^{2}, n(n+1)\right)$ and that the coordinates of $Q_{n+1}$ are $\left(((n+1)-1)^{2}, 0\right)=\left(n^{2}, 0\right)$.

Since $P_{n}$ and $Q_{n}$ have the same $x$-coordinate, then $P_{n} Q_{n}$ is vertical.

Since $P_{n+1}$ and $Q_{n+1}$ have the same $x$-coordinate, then $P_{n+1} Q_{n+1}$ is vertical.

Since $Q_{n}$ and $Q_{n+1}$ have the same $y$-coordinate, then $Q_{n} Q_{n+1}$ is horizontal.

![](https://cdn.mathpix.com/cropped/2024_04_30_75c1cf70fbe3735bfd92g-11.jpg?height=588&width=642&top_left_y=1175&top_left_x=796)

Let $A_{n}$ be the area of trapezoid $Q_{n} P_{n} P_{n+1} Q_{n+1}$.

Then

$$
\begin{aligned}
A_{n} & =\frac{1}{2}\left(P_{n} Q_{n}+P_{n+1} Q_{n+1}\right) Q_{n} Q_{n+1} \\
& =\frac{1}{2}((n(n-1)-0)+(n(n+1)-0))\left(n^{2}-(n-1)^{2}\right) \\
& =\frac{1}{2}\left(2 n^{2}\right)(2 n-1) \\
& =n^{2}(2 n-1)
\end{aligned}
$$

We want to determine the number of integers $n$ with $2 \leq n \leq 99$ for which $A_{n}$ is a perfect square.

Since $n^{2}$ is a perfect square, then $A_{n}$ is a perfect square if and only if $2 n-1$ is a perfect square. We note that $2 n-1$ is odd and, since $2 \leq n \leq 99$, then $3 \leq 2 n-1 \leq 197$.

Thus, the possible perfect square values of $2 n-1$ are the odd perfect squares between 3 and 197 , which are $3^{2}, 5^{2}, 7^{2}, 9^{2}, 11^{2}, 13^{2}$ (that is, $9,25,49,81,121,169$ ).

Therefore, there are 6 integers $n$ that satisfy the given conditions.

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