Find all triplets $(a,b,p)$ where $a,b$ are positive integers and $p$ is a prime number satisfying: $\frac{1}{p}=\frac{1}{a^2}+\frac{1}{b^2}$
Given the equation:
\[
\frac{1}{p} = \frac{1}{a^2} + \frac{1}{b^2}
\]
we start by finding a common denominator for the right-hand side:
\[
\frac{1}{p} = \frac{a^2 + b^2}{a^2 b^2}
\]
Multiplying both sides by \(a^2 b^2 p\) to clear the denominators, we get:
\[
a^2 b^2 = p(a^2 + b^2)
\]
Rearranging terms, we have:
\[
a^2 b^2 - p a^2 - p b^2 = 0
\]
We can rewrite this as:
\[
p(a^2 + b^2) = a^2 b^2
\]
This can be factored as:
\[
(p - a^2)(p - b^2) = p^2
\]
We now consider the possible values for \((p - a^2)\) and \((p - b^2)\). Since \(p\) is a prime number, the factors of \(p^2\) are \(1, p, p^2\) and their negatives. We analyze each case:

1. \(p - a^2 = 1\) and \(p - b^2 = p^2\):
   \[
   p - a^2 = 1 \implies a^2 = p - 1
   \]
   \[
   p - b^2 = p^2 \implies b^2 = p - p^2
   \]
   Since \(b^2\) must be positive, this case is not possible.

2. \(p - a^2 = -1\) and \(p - b^2 = -p^2\):
   \[
   p - a^2 = -1 \implies a^2 = p + 1
   \]
   \[
   p - b^2 = -p^2 \implies b^2 = p + p^2
   \]
   Since \(b^2\) must be a positive integer, this case is not possible.

3. \(p - a^2 = p\) and \(p - b^2 = p\):
   \[
   p - a^2 = p \implies a^2 = 0
   \]
   \[
   p - b^2 = p \implies b^2 = 0
   \]
   Since \(a\) and \(b\) are positive integers, this case is not possible.

4. \(p - a^2 = -p\) and \(p - b^2 = -p\):
   \[
   p - a^2 = -p \implies a^2 = 2p
   \]
   \[
   p - b^2 = -p \implies b^2 = 2p
   \]
   Since \(a\) and \(b\) are positive integers, this case is not possible.

5. \(p - a^2 = -1\) and \(p - b^2 = -p\):
   \[
   p - a^2 = -1 \implies a^2 = p + 1
   \]
   \[
   p - b^2 = -p \implies b^2 = 2p
   \]
   Since \(a\) and \(b\) are positive integers, this case is not possible.

6. \(p - a^2 = -p\) and \(p - b^2 = -1\):
   \[
   p - a^2 = -p \implies a^2 = 2p
   \]
   \[
   p - b^2 = -1 \implies b^2 = p + 1
   \]
   Since \(a\) and \(b\) are positive integers, this case is not possible.

The only valid case is when \(a = b = p = 2\):
\[
\frac{1}{2} = \frac{1}{2^2} + \frac{1}{2^2} = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}
\]

The final answer is \( \boxed{ (2, 2, 2) } \)