Real numbers $x,y,a$ satisfy the equations $x+y = x^3 +y^3 = x^5 +y^5 = a$.
Find all possible values of $a$.
Given the equations:
\[ x + y = a \]
\[ x^3 + y^3 = a \]
\[ x^5 + y^5 = a \]

We need to find all possible values of \(a\).

1. **Case 1: \(a = 0\)**
   - If \(a = 0\), then \(x + y = 0\). This implies \(y = -x\).
   - Substituting \(y = -x\) into the other equations:
     \[ x^3 + (-x)^3 = 0 \implies x^3 - x^3 = 0 \]
     \[ x^5 + (-x)^5 = 0 \implies x^5 - x^5 = 0 \]
   - Both equations are satisfied, so \(a = 0\) is a valid solution.

2. **Case 2: \(a \neq 0\)**
   - Given \(x + y = a\), we can express \(y\) as \(y = a - x\).
   - Substituting \(y = a - x\) into the second equation:
     \[ x^3 + (a - x)^3 = a \]
     \[ x^3 + (a^3 - 3a^2x + 3ax^2 - x^3) = a \]
     \[ a^3 - 3a^2x + 3ax^2 = a \]
     \[ 3ax^2 - 3a^2x + a^3 = a \]
     \[ 3ax^2 - 3a^2x + a^3 - a = 0 \]
     \[ a(3x^2 - 3ax + a^2 - 1) = 0 \]
   - Since \(a \neq 0\), we can divide by \(a\):
     \[ 3x^2 - 3ax + a^2 - 1 = 0 \]
   - This is a quadratic equation in \(x\). Solving for \(x\):
     \[ x = \frac{3a \pm \sqrt{(3a)^2 - 4 \cdot 3 \cdot (a^2 - 1)}}{2 \cdot 3} \]
     \[ x = \frac{3a \pm \sqrt{9a^2 - 12a^2 + 12}}{6} \]
     \[ x = \frac{3a \pm \sqrt{-3a^2 + 12}}{6} \]
     \[ x = \frac{3a \pm \sqrt{12 - 3a^2}}{6} \]
     \[ x = \frac{3a \pm \sqrt{3(4 - a^2)}}{6} \]
     \[ x = \frac{3a \pm \sqrt{3} \sqrt{4 - a^2}}{6} \]
     \[ x = \frac{a \pm \sqrt{4 - a^2}}{2} \]
   - For \(x\) to be real, the discriminant must be non-negative:
     \[ 4 - a^2 \geq 0 \]
     \[ a^2 \leq 4 \]
     \[ -2 \leq a \leq 2 \]

3. **Verification of possible values of \(a\)**
   - We need to check if the values \(a = \pm 1\) and \(a = \pm 2\) satisfy the original equations.
   - For \(a = \pm 1\):
     \[ x + y = \pm 1 \]
     \[ x^3 + y^3 = \pm 1 \]
     \[ x^5 + y^5 = \pm 1 \]
   - For \(a = \pm 2\):
     \[ x + y = \pm 2 \]
     \[ x^3 + y^3 = \pm 2 \]
     \[ x^5 + y^5 = \pm 2 \]

   - These values satisfy the original equations, confirming that \(a = \pm 1\) and \(a = \pm 2\) are valid solutions.

The final answer is \(\boxed{a = 0, \pm 1, \pm 2}\).