a) Find all the integers $m$ and $n$ such that

\[9m^2+3n=n^2+8\]

b) Let $a,b\in \mathbb{N}^*$ . If $x=a^{a+b}+(a+b)^a$ and $y=a^a+(a+b)^{a+b}$ which one is bigger?

[i]Florin Nicoara, Valer Pop[/i]
### Part (a)

We start with the given equation:
\[ 9m^2 + 3n = n^2 + 8 \]

1. Rearrange the equation to form a quadratic in \( n \):
\[ n^2 - 3n + (8 - 9m^2) = 0 \]

2. For \( n \) to be an integer, the discriminant of this quadratic equation must be a perfect square. The discriminant \(\Delta\) of the quadratic equation \( n^2 - 3n + (8 - 9m^2) = 0 \) is given by:
\[ \Delta = b^2 - 4ac = (-3)^2 - 4 \cdot 1 \cdot (8 - 9m^2) \]
\[ \Delta = 9 - 4(8 - 9m^2) \]
\[ \Delta = 9 - 32 + 36m^2 \]
\[ \Delta = 36m^2 - 23 \]

3. For \( n \) to be an integer, \(\Delta\) must be a perfect square. Let \(\Delta = k^2\) for some integer \( k \):
\[ 36m^2 - 23 = k^2 \]

4. This is a Diophantine equation. We need to find integer solutions for \( m \) and \( k \). Rearrange the equation:
\[ 36m^2 - k^2 = 23 \]
\[ (6m - k)(6m + k) = 23 \]

5. The factors of 23 are \( (1, 23) \) and \( (-1, -23) \). We solve for \( m \) and \( k \) using these pairs:
   - For \( 6m - k = 1 \) and \( 6m + k = 23 \):
     \[ 6m - k = 1 \]
     \[ 6m + k = 23 \]
     Adding these equations:
     \[ 12m = 24 \]
     \[ m = 2 \]
     Substituting \( m = 2 \) back into \( 6m - k = 1 \):
     \[ 6(2) - k = 1 \]
     \[ 12 - k = 1 \]
     \[ k = 11 \]

   - For \( 6m - k = -1 \) and \( 6m + k = -23 \):
     \[ 6m - k = -1 \]
     \[ 6m + k = -23 \]
     Adding these equations:
     \[ 12m = -24 \]
     \[ m = -2 \]
     Substituting \( m = -2 \) back into \( 6m - k = -1 \):
     \[ 6(-2) - k = -1 \]
     \[ -12 - k = -1 \]
     \[ k = -11 \]

6. Now, substitute \( m = 2 \) and \( m = -2 \) back into the quadratic equation to find \( n \):
   - For \( m = 2 \):
     \[ n^2 - 3n + (8 - 9(2)^2) = 0 \]
     \[ n^2 - 3n + (8 - 36) = 0 \]
     \[ n^2 - 3n - 28 = 0 \]
     Solving this quadratic equation:
     \[ n = \frac{3 \pm \sqrt{9 + 112}}{2} \]
     \[ n = \frac{3 \pm 11}{2} \]
     \[ n = 7 \text{ or } n = -4 \]

   - For \( m = -2 \):
     \[ n^2 - 3n + (8 - 9(-2)^2) = 0 \]
     \[ n^2 - 3n + (8 - 36) = 0 \]
     \[ n^2 - 3n - 28 = 0 \]
     Solving this quadratic equation:
     \[ n = \frac{3 \pm \sqrt{9 + 112}}{2} \]
     \[ n = \frac{3 \pm 11}{2} \]
     \[ n = 7 \text{ or } n = -4 \]

Thus, the integer solutions are:
\[ (m, n) = (2, 7), (2, -4), (-2, 7), (-2, -4) \]

### Part (b)

Given:
\[ x = a^{a+b} + (a+b)^a \]
\[ y = a^a + (a+b)^{a+b} \]

1. Consider the function \( f(n) = n^{a+b} - n^a \). We need to compare \( f(a) \) and \( f(a+b) \):
\[ f(a) = a^{a+b} - a^a \]
\[ f(a+b) = (a+b)^{a+b} - (a+b)^a \]

2. Since \( a \) and \( b \) are natural numbers, \( a < a+b \). We need to show that \( f(n) \) is strictly increasing for natural numbers:
\[ f(n) = n^a (n^b - 1) \]
Since \( n^a \) and \( n^b - 1 \) are both positive and increasing for \( n \in \mathbb{N}^* \), \( f(n) \) is strictly increasing.

3. Therefore, \( f(a) < f(a+b) \):
\[ a^{a+b} - a^a < (a+b)^{a+b} - (a+b)^a \]
\[ a^{a+b} + (a+b)^a < (a+b)^{a+b} + a^a \]
\[ x < y \]

The final answer is \( \boxed{ x < y } \)