imc

If $P(n)=\sqrt{n}$, then $n=p_1^2$, where $p_1$ is a prime number. If $P(n+48)=\sqrt{n+48}$, then $n+48$ is a square, but we know that n is $p_1^2$. This means we just have to check for squares of primes, add $48$ and look whether the root is a prime number. We can easily see that the difference between two consecutive square after $576$ is greater than or equal to $49$, Hence we have to consider only the prime numbers till $23$. Squaring prime numbers below $23$ including $23$ we get the following list. $4,9,25,49,121,169,289,361,529$ But adding $48$ to a number ending with $9$ will result in a number ending with $7$, but we know that a perfect square does not end in $7$, so we can eliminate those cases to get the new list. $4,25,121,361$ Adding $48$, we get $121$ as the only possible solution. Hence the answer is $\boxed{1}$.