IRL ABooklet

Solution 1. We claim that $q(n) = 2^n - 4$ and that this is a perfect square only when $n = 3$.
If there is a unique turning point, then it is either a minimum or maximum. We count the number of permutations where the turning point is a maximum, and so $q(n)$ is double this number. As we are interested in the permutations with a single turning point, the permutation must be increasing to the left of the maximum and decreasing to its right. So a permutation with a unique maximum is fully determined by the way the remaining numbers are allocated to the left or the right of the maximum.
The maximal value must be $p_t = n$. Consider the $n-1$ numbers $1, 2, \dots, n-1$. These must each be placed either to the left of the turning point $t$ or to the right of $t$. There are $2^{n-1}$ ways of allocating the numbers between left and right sides, but as a turning point $t$ must (according to our definition) be an interior point, we must exclude the 2 cases where all the numbers from 1 to $n-1$ are placed the same side of $t$. This leaves $2^{n-1} - 2$ ways of allocating the points between left and right sides.
Doubling, it follows that there are $2^n - 4$ permutations with exactly one turning point, either a minimum or a maximum.
Now for $n \ge 3$, $2^n - 4$ is an even number, so is a perfect square, say $(2r)^2$, if and only if a quarter of that number, namely $2^{n-2} - 1 = r^2$ is a perfect square. If $n \ge 4$ then $2^{n-2}$ is a multiple of 4, so $2^{n-2} - 1 \equiv 3 \pmod 4$ and cannot be a perfect square. Therefore, the only $n \ge 3$ where $q(n) = 2^n - 4$ is a perfect square is $q(3) = 2^3 - 4 = 4$.

Solution 2. An alternative calculation of $q(n)$ considers that if $p_t = n$ then there are $\binom{n-1}{t-1}$ ways to allocate the remaining $n-1$ numbers such that $t-1$ are to the left of $p_t$ and $n-t$ are to the right. The total number of cases for which $p_t$ is a maximum is then
$$\frac{q(n)}{2} = \sum_{t=2}^{n-1} \binom{n-1}{t-1}$$
The binomial theorem gives
$$2^{n-1} = (1+1)^{n-1} = \sum_{t=1}^{n} \binom{n-1}{t-1} = 2 + \sum_{t=2}^{n-1} \binom{n-1}{t-1}$$
The enumeration $q(n) = 2^n - 4$ follows. Finish as in Solution 1.

Solution 3. If a permutation has only one turning point, that point has to be 1 or $n$. There is the same number of permutations in both cases. We find a recurrence relation for $q(n)$:
$$q(n + 1) = 2q(n) + 4 \quad \text{for } n \ge 3.$$
Indeed, let $\sigma$ be a permutation on $\{1, 2, \dots, n\}$. If $\sigma$ has 1 as unique turning point, then we can add $n+1$ either at the beginning or the end to get a permutation $\tilde{\sigma}$ with 1 turning point. If $\sigma$ has $n$ as unique turning point, then we can add $n+1$ either at immediately left or immediately right of $n$ to get a permutation $\tilde{\sigma}$ with $n+1$ as unique turning point. If $\sigma$ is strictly increasing, then add $n+1$ either at the beginning or just left of $n$ and if $\sigma$ is strictly decreasing, then add $n+1$ either at the end or just right of $n$. These are all possible permutations of $\{1, 2, \dots, n, n+1\}$ with 1 turning point.
To prove that $n=3$ is the unique number for which $q(n) = m^2$, we first note that $q(3) = 4$, because there are only four permutations of $\{1, 2, 3\}$ with exactly one turning point, namely
$$(2, 1, 3) \quad (3, 1, 2) \quad (1, 3, 2) \quad (2, 3, 1).$$
Using induction and the recurrence relation we see that $4 \mid q(n)$ for all $n \ge 3$. Writing $q(n) = 4x_n$ we get $x_{n+1} = 2x_n + 1$ for all $n \ge 3$ from the recurrence relation for $q$. If $x_{n+1} = k^2$ is a square for some $n \ge 3$, $k$ must be odd, hence $k^2 \equiv 1 \pmod 4$. Since $k^2 = x_{n+1} = 2x_n + 1$, we have $2x_n = k^2 - 1 \equiv 0 \pmod 4$, hence $2 \mid x_n$. When $n=3$, this contradicts $x_3 = 1$, and when $n \ge 4$ this contradicts $x_n = 2x_{n-1} + 1$ which holds for $n > 3$.