Determine all integers $a$ for which the equation \[x^{2}+axy+y^{2}=1\] has infinitely many distinct integer solutions $x, \;y$.
To determine all integers \( a \) for which the equation 
\[ x^2 + axy + y^2 = 1 \]
has infinitely many distinct integer solutions \( x \) and \( y \), we need to analyze the given equation and find conditions on \( a \).

1. **Rewrite the equation**:
   \[ x^2 + axy + y^2 = 1 \]

2. **Consider the discriminant**:
   For the quadratic equation in \( x \):
   \[ x^2 + axy + y^2 - 1 = 0 \]
   The discriminant of this quadratic equation in \( x \) must be a perfect square for \( x \) to be an integer. The discriminant is given by:
   \[ \Delta = (ay)^2 - 4(y^2 - 1) = a^2y^2 - 4y^2 + 4 = (a^2 - 4)y^2 + 4 \]

3. **Perfect square condition**:
   For \( x \) to be an integer, \( \Delta \) must be a perfect square. Let \( k \) be an integer such that:
   \[ (a^2 - 4)y^2 + 4 = k^2 \]
   Rearrange this equation:
   \[ (a^2 - 4)y^2 = k^2 - 4 \]
   \[ (a^2 - 4)y^2 = (k - 2)(k + 2) \]

4. **Analyze the factors**:
   For the equation to have infinitely many solutions, \( (a^2 - 4) \) must be a perfect square. Let \( a^2 - 4 = m^2 \) for some integer \( m \):
   \[ a^2 - 4 = m^2 \]
   \[ a^2 - m^2 = 4 \]
   \[ (a - m)(a + m) = 4 \]

5. **Solve the factor pairs**:
   The factor pairs of 4 are:
   \[ (1, 4), (-1, -4), (2, 2), (-2, -2) \]
   Solving these pairs:
   - \( a - m = 1 \) and \( a + m = 4 \):
     \[ 2a = 5 \Rightarrow a = \frac{5}{2} \] (not an integer)
   - \( a - m = -1 \) and \( a + m = -4 \):
     \[ 2a = -5 \Rightarrow a = -\frac{5}{2} \] (not an integer)
   - \( a - m = 2 \) and \( a + m = 2 \):
     \[ 2a = 4 \Rightarrow a = 2 \]
   - \( a - m = -2 \) and \( a + m = -2 \):
     \[ 2a = -4 \Rightarrow a = -2 \]

6. **Verify the solutions**:
   For \( a = 2 \):
   \[ x^2 + 2xy + y^2 = 1 \]
   This equation can be rewritten as:
   \[ (x + y)^2 - xy = 1 \]
   This equation has infinitely many integer solutions, for example, \( (1, -1) \), \( (2, -3) \), etc.

   For \( a = -2 \):
   \[ x^2 - 2xy + y^2 = 1 \]
   This equation can be rewritten as:
   \[ (x - y)^2 + xy = 1 \]
   This equation also has infinitely many integer solutions, for example, \( (1, 1) \), \( (2, 3) \), etc.

Conclusion:
The integers \( a \) for which the equation \( x^2 + axy + y^2 = 1 \) has infinitely many distinct integer solutions are \( a = 2 \) and \( a = -2 \).

The final answer is \( \boxed{ a = 2 } \) and \( a = -2 \).