Find all real $x, y$ such that  $\begin{cases}(1 + x)(1 + x^2)(1 + x^4) = 1+ y^7 \\
 (1 + y)(1 + y^2)(1 + y^4) = 1+ x^7 \end{cases}$ .
1. We start with the given system of equations:
   \[
   \begin{cases}
   (1 + x)(1 + x^2)(1 + x^4) = 1 + y^7 \\
   (1 + y)(1 + y^2)(1 + y^4) = 1 + x^7
   \end{cases}
   \]

2. Without loss of generality (WLOG), assume \( x \geq y \). This implies:
   \[
   1 + x^7 \geq 1 + y^7 \implies x^7 \geq y^7
   \]
   Since \( x \geq y \), we also have:
   \[
   (1 + y)(1 + y^2)(1 + y^4) \geq (1 + x)(1 + x^2)(1 + x^4)
   \]

3. From the above inequalities, we get:
   \[
   y \geq x
   \]
   Combining \( x \geq y \) and \( y \geq x \), we conclude:
   \[
   x = y
   \]

4. Substituting \( x = y \) into either of the original equations, we get:
   \[
   (1 + x)(1 + x^2)(1 + x^4) = 1 + x^7
   \]

5. Expanding the left-hand side:
   \[
   (1 + x)(1 + x^2)(1 + x^4) = 1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7
   \]

6. Equating this to the right-hand side:
   \[
   1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 = 1 + x^7
   \]

7. Simplifying, we get:
   \[
   x + x^2 + x^3 + x^4 + x^5 + x^6 = 0
   \]

8. Factoring out \( x \):
   \[
   x(x + x^2 + x^3 + x^4 + x^5 + 1) = 0
   \]

9. This gives us two cases:
   \[
   x = 0 \quad \text{or} \quad x^6 + x^4 + x^2 + 1 = 0
   \]

10. The polynomial \( x^6 + x^4 + x^2 + 1 \) can be rewritten as:
    \[
    x^6 + x^4 + x^2 + 1 = (x^2 + 1)(x^4 + 1)
    \]
    Since \( x^2 + 1 \geq 1 \) and \( x^4 + 1 \geq 1 \) for all real \( x \), there are no real solutions for \( x^6 + x^4 + x^2 + 1 = 0 \).

11. Therefore, the only real solutions are:
    \[
    x = 0 \quad \text{or} \quad x = -1
    \]

12. Since \( x = y \), we have:
    \[
    y = 0 \quad \text{or} \quad y = -1
    \]

The final answer is \( \boxed{x = y = 0 \text{ or } x = y = -1} \).