Let $P(x) = x^2-1$ be a polynomial, and let $a$ be a positive real number satisfying$$P(P(P(a))) = 99.$$ The value of $a^2$ can be written as $m+\sqrt{n}$, where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$.

[i]Proposed by [b]HrishiP[/b][/i]
1. First, we start with the polynomial \( P(x) = x^2 - 1 \). We need to find \( a \) such that \( P(P(P(a))) = 99 \).

2. Calculate \( P(P(a)) \):
   \[
   P(a) = a^2 - 1
   \]
   \[
   P(P(a)) = P(a^2 - 1) = (a^2 - 1)^2 - 1 = a^4 - 2a^2 + 1 - 1 = a^4 - 2a^2
   \]

3. Calculate \( P(P(P(a))) \):
   \[
   P(P(P(a))) = P(a^4 - 2a^2) = (a^4 - 2a^2)^2 - 1
   \]
   \[
   (a^4 - 2a^2)^2 - 1 = (a^4 - 2a^2 + 1)(a^4 - 2a^2 - 1)
   \]

4. Set the expression equal to 99:
   \[
   (a^4 - 2a^2)^2 - 1 = 99
   \]
   \[
   (a^4 - 2a^2)^2 = 100
   \]
   \[
   a^4 - 2a^2 = \pm 10
   \]

5. Solve the quadratic equation \( a^4 - 2a^2 = 10 \) and \( a^4 - 2a^2 = -10 \):
   - For \( a^4 - 2a^2 = 10 \):
     \[
     a^4 - 2a^2 - 10 = 0
     \]
     Let \( y = a^2 \):
     \[
     y^2 - 2y - 10 = 0
     \]
     Solve using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
     \[
     y = \frac{2 \pm \sqrt{4 + 40}}{2} = \frac{2 \pm \sqrt{44}}{2} = 1 \pm \sqrt{11}
     \]
     Since \( y = a^2 \) and \( y \) must be positive:
     \[
     a^2 = 1 + \sqrt{11}
     \]

   - For \( a^4 - 2a^2 = -10 \):
     \[
     a^4 - 2a^2 + 10 = 0
     \]
     Let \( y = a^2 \):
     \[
     y^2 - 2y + 10 = 0
     \]
     The discriminant \( b^2 - 4ac = 4 - 40 = -36 \) is negative, so there are no real solutions.

6. Therefore, the only valid solution is \( a^2 = 1 + \sqrt{11} \).

7. The value of \( a^2 \) can be written as \( m + \sqrt{n} \) where \( m = 1 \) and \( n = 11 \).

8. Thus, \( m + n = 1 + 11 = 12 \).

The final answer is \( \boxed{12} \).