Let $n>1$ be a natural number. Find the real values of the parameter $a$, for which the equation $\sqrt[n]{1+x}+\sqrt[n]{1-x}=a$ has a single real root.
1. Consider the given equation:
   \[
   \sqrt[n]{1+x} + \sqrt[n]{1-x} = a
   \]
   where \( n > 1 \) is a natural number and \( a \) is a real parameter.

2. Define the function:
   \[
   f(x) = \sqrt[n]{1+x} + \sqrt[n]{1-x}
   \]
   We need to analyze the behavior of \( f(x) \) to determine for which values of \( a \) the equation \( f(x) = a \) has a single real root.

3. First, observe that \( f(x) \) is symmetric around \( x = 0 \). This is because:
   \[
   f(-x) = \sqrt[n]{1-x} + \sqrt[n]{1+x} = f(x)
   \]
   Therefore, \( f(x) \) is an even function.

4. Evaluate \( f(x) \) at \( x = 0 \):
   \[
   f(0) = \sqrt[n]{1+0} + \sqrt[n]{1-0} = 1 + 1 = 2
   \]
   Thus, \( f(0) = 2 \).

5. Next, consider the behavior of \( f(x) \) as \( x \) approaches the boundaries of its domain, which is \( -1 \leq x \leq 1 \):
   - As \( x \to 1 \):
     \[
     \sqrt[n]{1+x} \to \sqrt[n]{2}, \quad \sqrt[n]{1-x} \to \sqrt[n]{0} = 0
     \]
     Hence, \( f(1) = \sqrt[n]{2} + 0 = \sqrt[n]{2} \).
   - As \( x \to -1 \):
     \[
     \sqrt[n]{1+x} \to \sqrt[n]{0} = 0, \quad \sqrt[n]{1-x} \to \sqrt[n]{2}
     \]
     Hence, \( f(-1) = 0 + \sqrt[n]{2} = \sqrt[n]{2} \).

6. Since \( f(x) \) is continuous and symmetric, and \( f(0) = 2 \) is the maximum value, we have:
   \[
   f(x) \leq 2 \quad \text{for all} \quad -1 \leq x \leq 1
   \]

7. To determine if \( f(x) = 2 \) has only one real root, we need to check the derivative \( f'(x) \):
   \[
   f'(x) = \frac{d}{dx} \left( \sqrt[n]{1+x} + \sqrt[n]{1-x} \right) = \frac{1}{n} (1+x)^{\frac{1}{n}-1} - \frac{1}{n} (1-x)^{\frac{1}{n}-1}
   \]
   At \( x = 0 \):
   \[
   f'(0) = \frac{1}{n} (1+0)^{\frac{1}{n}-1} - \frac{1}{n} (1-0)^{\frac{1}{n}-1} = 0
   \]
   Since \( f(x) \) is symmetric and \( f'(0) = 0 \), \( x = 0 \) is a critical point.

8. To confirm that \( x = 0 \) is the only root when \( a = 2 \), we need to check the second derivative \( f''(x) \):
   \[
   f''(x) = \frac{d}{dx} \left( \frac{1}{n} (1+x)^{\frac{1}{n}-1} - \frac{1}{n} (1-x)^{\frac{1}{n}-1} \right)
   \]
   At \( x = 0 \):
   \[
   f''(0) = \frac{1}{n} \left( \frac{1}{n} - 1 \right) (1+0)^{\frac{1}{n}-2} + \frac{1}{n} \left( \frac{1}{n} - 1 \right) (1-0)^{\frac{1}{n}-2} = \frac{2}{n} \left( \frac{1}{n} - 1 \right)
   \]
   Since \( n > 1 \), \( \frac{1}{n} - 1 < 0 \), thus \( f''(0) < 0 \), indicating that \( x = 0 \) is a local maximum.

9. Therefore, \( f(x) = 2 \) has only one real root at \( x = 0 \) when \( a = 2 \).

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