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 $n+48 = p_{2}^{2}$, where $p_{2}$ is a different prime number. So: $p_{2}^{2} = n+48$ $p_{1}^{2} = n$ $p_{2}^{2} - p_{1}^{2} = 48$ $(p_{2}+p_{1})(p_{2}-p_{1})=48$ Since $p_{1} > 0$: $(p_{2}+p_{1}) > (p_{2}-p_{1})$. Looking at pairs of divisors of $48$, we have several possibilities to solve for $p_{1}$ and $p_{2}$: $(p_{2}+p_{1}) = 48$ $(p_{2}-p_{1}) = 1$ $p_{1} = \frac{47}{2}$ $p_{2} = \frac{49}{2}$ $(p_{2}+p_{1}) = 24$ $(p_{2}-p_{1}) = 2$ $p_{1} = 11$ $p_{2} = 13$ $(p_{2}+p_{1}) = 16$ $(p_{2}-p_{1}) = 3$ $p_{1} = \frac{13}{2}$ $p_{2} = \frac{19}{2}$ $(p_{2}+p_{1}) = 12$ $(p_{2}-p_{1}) = 4$ $p_{1} = 4$ $p_{2} = 8$ $(p_{2}+p_{1}) = 8$ $(p_{2}-p_{1}) = 6$ $p_{1} = 1$ $p_{2} = 7$ The only solution $(p_{1} , p_{2})$ where both numbers are primes is $(11,13)$. Therefore the number of positive integers $n$ that satisfy both statements is $1$.