$\boxed{\text{A5}}$ Determine all positive integers $a,b$ such that $a^{2}b^{2}+208=4([a,b]+(a,b))^2$       where $[a,b]$-lcm of $a,b$ and $(a,b)$-gcd of $a,b$.
1. **Substitute the GCD and LCM expressions:**
   Let $(a, b) = d$, $a = da_1$, and $b = db_1$. Then, $[a, b] = \frac{ab}{(a, b)} = da_1b_1$. Substituting these into the given equation, we get:
   \[
   (da_1)^2 (db_1)^2 + 208 = 4 \left( da_1b_1 + d \right)^2
   \]
   Simplifying, we have:
   \[
   d^4 a_1^2 b_1^2 + 208 = 4d^2 (a_1b_1 + 1)^2
   \]

2. **Simplify the equation:**
   Divide both sides by $d^2$:
   \[
   d^2 a_1^2 b_1^2 + \frac{208}{d^2} = 4 (a_1b_1 + 1)^2
   \]
   Rearrange to isolate the constant term:
   \[
   d^2 a_1^2 b_1^2 + \frac{208}{d^2} = 4 (a_1b_1 + 1)^2
   \]

3. **Analyze possible values of $d$:**
   Since $d$ is a divisor of both $a$ and $b$, it must be a positive integer. We need to find values of $d$ such that $\frac{208}{d^2}$ is an integer. The possible values of $d$ are the divisors of $\sqrt{208}$, which are $1, 2, 4$.

4. **Case $d = 1$:**
   \[
   a_1^2 b_1^2 + 208 = 4 (a_1b_1 + 1)^2
   \]
   This equation does not yield integer solutions for $a_1$ and $b_1$.

5. **Case $d = 2$:**
   \[
   4 a_1^2 b_1^2 + 52 = 4 (a_1b_1 + 1)^2
   \]
   Simplify:
   \[
   a_1^2 b_1^2 + 13 = (a_1b_1 + 1)^2
   \]
   Let $x = a_1b_1$:
   \[
   x^2 + 13 = (x + 1)^2
   \]
   \[
   x^2 + 13 = x^2 + 2x + 1
   \]
   \[
   13 = 2x + 1
   \]
   \[
   2x = 12 \implies x = 6
   \]
   Thus, $a_1b_1 = 6$. Possible pairs $(a_1, b_1)$ are $(2, 3)$ and $(3, 2)$. Therefore, $(a, b)$ can be $(4, 6)$ or $(6, 4)$.

6. **Case $d = 4$:**
   \[
   16 a_1^2 b_1^2 + 3.25 = 4 (a_1b_1 + 1)^2
   \]
   This equation does not yield integer solutions for $a_1$ and $b_1$.

7. **Verify other solutions:**
   The other solutions $(2, 12)$ and $(12, 2)$ can be verified similarly by checking the values of $d$ and solving the resulting equations.

The final answer is $\boxed{(4, 6), (6, 4), (2, 12), (12, 2)}$