Find all real numbers a such that all solutions to the quadratic equation $ x^2 \minus{} ax \plus{} a \equal{} 0$ are integers.
1. Let \( m \) and \( n \) be the roots of the quadratic equation \( x^2 - ax + a = 0 \). By Vieta's formulas, we know:
   \[
   m + n = a \quad \text{and} \quad mn = a
   \]
   Since \( m \) and \( n \) are integers, \( a \) must also be an integer.

2. The roots of the quadratic equation can be expressed using the quadratic formula:
   \[
   m = \frac{a + \sqrt{a^2 - 4a}}{2}, \quad n = \frac{a - \sqrt{a^2 - 4a}}{2}
   \]
   Since \( m \) and \( n \) are integers, \( \sqrt{a^2 - 4a} \) must be an integer. Let \( \sqrt{a^2 - 4a} = k \), where \( k \) is a positive integer.

3. Therefore, we have:
   \[
   a^2 - 4a = k^2
   \]
   Rearrange this equation to:
   \[
   a^2 - 4a - k^2 = 0
   \]
   This can be factored as:
   \[
   (a - 2)^2 - k^2 = 4
   \]
   Using the difference of squares, we get:
   \[
   (a - 2 + k)(a - 2 - k) = 4
   \]

4. Since \( a \) and \( k \) are integers, \( (a - 2 + k) \) and \( (a - 2 - k) \) must be pairs of factors of 4. The possible pairs of factors of 4 are:
   \[
   (1, 4), (-1, -4), (2, 2), (-2, -2)
   \]

5. Solve for \( a \) and \( k \) for each pair:
   - For \( (a - 2 + k) = 4 \) and \( (a - 2 - k) = 1 \):
     \[
     a - 2 + k = 4 \quad \text{and} \quad a - 2 - k = 1
     \]
     Adding these equations:
     \[
     2(a - 2) = 5 \implies a - 2 = \frac{5}{2} \implies a = \frac{9}{2} \quad \text{(not an integer)}
     \]
   - For \( (a - 2 + k) = -1 \) and \( (a - 2 - k) = -4 \):
     \[
     a - 2 + k = -1 \quad \text{and} \quad a - 2 - k = -4
     \]
     Adding these equations:
     \[
     2(a - 2) = -5 \implies a - 2 = -\frac{5}{2} \implies a = -\frac{1}{2} \quad \text{(not an integer)}
     \]
   - For \( (a - 2 + k) = 2 \) and \( (a - 2 - k) = 2 \):
     \[
     a - 2 + k = 2 \quad \text{and} \quad a - 2 - k = 2
     \]
     Subtracting these equations:
     \[
     2k = 0 \implies k = 0
     \]
     Then:
     \[
     a - 2 = 2 \implies a = 4
     \]
   - For \( (a - 2 + k) = -2 \) and \( (a - 2 - k) = -2 \):
     \[
     a - 2 + k = -2 \quad \text{and} \quad a - 2 - k = -2
     \]
     Subtracting these equations:
     \[
     2k = 0 \implies k = 0
     \]
     Then:
     \[
     a - 2 = -2 \implies a = 0
     \]

6. Therefore, the possible values of \( a \) are \( 0 \) and \( 4 \).

The final answer is \( \boxed{ a = 0, 4 } \)