### Let $p$ be a prime number such that the next larger number is a perfect square. Find the sum of all such prime numbers. $($For example, if you think $11$ and $13$ are two such prime numbers, then the sum is $24.)$

$3$

Step by Step Explanation:
1. Given, $p$ is a prime number such that the next larger number is a perfect square.
$i.e., p + 1 = n^2$
2. Rewriting $p + 1 = n^2,$ we get
\begin{align} & p = n^2 - 1 \\ \implies & p = (n - 1)(n +1) \end{align}
3. Since $p$ is a prime number, therefore $n-1$ needs to be equal to $1.$
$\implies n = 2$
Therefore, $p = 3$ is unique.
Hence, the required sum is $3.$