The arithmetic mean of two different positive integers $x,y$ is a two digit integer. If one interchanges the digits, the geometric mean of these numbers is archieved.
a) Find $x,y$.
b) Show that a)'s solution is unique up to permutation if we work in base $g=10$, but that there is no solution in base $g=12$.
c) Give more numbers $g$ such that a) can be solved; give more of them such that a) can't be solved, too.
1. **Given Conditions and Definitions:**
   - The arithmetic mean of two different positive integers \( x \) and \( y \) is a two-digit integer.
   - Interchanging the digits of this two-digit integer gives the geometric mean of \( x \) and \( y \).

2. **Formulating the Problem:**
   - Let the arithmetic mean be \(\overline{ab}\), where \(a\) and \(b\) are digits.
   - Therefore, \(\frac{x + y}{2} = 10a + b\).
   - The geometric mean is \(\overline{ba}\), so \(\sqrt{xy} = 10b + a\).

3. **Deriving the Relationship:**
   - From the arithmetic mean, we have:
     \[
     x + y = 2(10a + b)
     \]
   - From the geometric mean, we have:
     \[
     xy = (10b + a)^2
     \]

4. **Solving for \(x\) and \(y\):**
   - We need to solve the system of equations:
     \[
     x + y = 20a + 2b
     \]
     \[
     xy = (10b + a)^2
     \]
   - Assume \(x \neq y\), hence \(\frac{x + y}{2} > \sqrt{xy}\), implying \(10a + b > 10b + a\), which simplifies to \(a > b\).

5. **Expressing \(x\) and \(y\) in Terms of \(a\) and \(b\):**
   - Let \(x\) and \(y\) be the roots of the quadratic equation:
     \[
     t^2 - (20a + 2b)t + (10b + a)^2 = 0
     \]
   - The roots are given by:
     \[
     t = \frac{(20a + 2b) \pm \sqrt{(20a + 2b)^2 - 4(10b + a)^2}}{2}
     \]
   - Simplifying the discriminant:
     \[
     (20a + 2b)^2 - 4(10b + a)^2 = 400a^2 + 80ab + 4b^2 - 400b^2 - 80ab - 4a^2 = 396a^2 - 396b^2 = 396(a^2 - b^2)
     \]
   - Therefore, the roots are:
     \[
     t = 10a + b \pm \sqrt{99(a^2 - b^2)}
     \]

6. **Ensuring Integer Solutions:**
   - For \(t\) to be an integer, \(\sqrt{99(a^2 - b^2)}\) must be an integer.
   - This implies \(99(a^2 - b^2)\) must be a perfect square.
   - Since \(99 = 3^2 \times 11\), \(a^2 - b^2\) must be divisible by 11.

7. **Possible Values for \(a\) and \(b\):**
   - Given \(a, b \in \{1, 2, \ldots, 9\}\), we need \(11 \mid (a + b)(a - b)\).
   - Possible pairs \((a, b)\) are:
     \[
     (6, 5), (7, 4), (8, 3), (9, 2)
     \]

8. **Checking Each Pair:**
   - For \((a, b) = (6, 5)\):
     \[
     x + y = 20 \cdot 6 + 2 \cdot 5 = 130
     \]
     \[
     xy = (10 \cdot 5 + 6)^2 = 56^2 = 3136
     \]
     - Solving \(t^2 - 130t + 3136 = 0\) gives \(x = 98\) and \(y = 32\).

9. **Uniqueness in Base 10:**
   - The solution \((x, y) = (98, 32)\) is unique up to permutation in base 10.

10. **No Solution in Base 12:**
    - In base 12, \(g = 12\):
      \[
      g^2 - 1 = 143 = 11 \times 13
      \]
    - Since \(11 \times 13 > 24\), there is no solution in base 12.

11. **Other Bases:**
    - For \(g = 9\), \(g^2 - 1 = 80\), which is less than \(18\), so a solution exists.
    - For \(g = 14\), \(g^2 - 1 = 195\), which is greater than \(28\), so no solution exists.

The final answer is \( \boxed{ (x, y) = (98, 32) } \) or \( (32, 98) \).