Find the solution of the functional equation $P(x)+P(1-x)=1$ with power $2015$

P.S: $P(y)=y^{2015}$ is also a function with power $2015$
1. We start with the given functional equation:
   \[
   P(x) + P(1-x) = 1
   \]
   We need to find a polynomial \( P(x) \) of degree 2015 that satisfies this equation.

2. Let's assume \( P(x) \) is a polynomial of degree 2015. We can write:
   \[
   P(x) = a_{2015} x^{2015} + a_{2014} x^{2014} + \cdots + a_1 x + a_0
   \]

3. Substitute \( x \) and \( 1-x \) into the functional equation:
   \[
   P(x) + P(1-x) = 1
   \]
   This implies:
   \[
   a_{2015} x^{2015} + a_{2014} x^{2014} + \cdots + a_1 x + a_0 + a_{2015} (1-x)^{2015} + a_{2014} (1-x)^{2014} + \cdots + a_1 (1-x) + a_0 = 1
   \]

4. Notice that \( (1-x)^{2015} \) can be expanded using the binomial theorem:
   \[
   (1-x)^{2015} = \sum_{k=0}^{2015} \binom{2015}{k} (-x)^k = \sum_{k=0}^{2015} \binom{2015}{k} (-1)^k x^k
   \]

5. Therefore, the equation becomes:
   \[
   a_{2015} x^{2015} + a_{2014} x^{2014} + \cdots + a_1 x + a_0 + a_{2015} \sum_{k=0}^{2015} \binom{2015}{k} (-1)^k x^k + a_{2014} \sum_{k=0}^{2014} \binom{2014}{k} (-1)^k x^k + \cdots + a_1 (1-x) + a_0 = 1
   \]

6. Grouping the terms with the same powers of \( x \), we get:
   \[
   \sum_{k=0}^{2015} \left( a_{2015} + a_{2015} (-1)^k \right) x^k + \sum_{k=0}^{2014} \left( a_{2014} + a_{2014} (-1)^k \right) x^k + \cdots + \left( a_1 + a_1 (-1) \right) x + 2a_0 = 1
   \]

7. For the equation to hold for all \( x \), the coefficients of each power of \( x \) must match. This implies:
   \[
   a_{2015} + a_{2015} (-1)^k = 0 \quad \text{for all odd } k
   \]
   \[
   a_{2014} + a_{2014} (-1)^k = 0 \quad \text{for all even } k
   \]
   \[
   \vdots
   \]
   \[
   a_1 + a_1 (-1) = 0
   \]
   \[
   2a_0 = 1
   \]

8. From the above, we see that \( a_{2015} \) must be zero for all odd \( k \), and similarly for all even \( k \). This means that all coefficients of even powers of \( x \) must vanish, and \( a_0 = \frac{1}{2} \).

9. Therefore, \( P(x) \) must be of the form:
   \[
   P(x) = Q\left(\frac{1}{2} - x\right) + \frac{1}{2}
   \]
   where \( Q \) is an odd polynomial of degree 2015.

10. To verify, if \( Q \) is any such polynomial of degree 2015, then:
    \[
    P(x) = Q\left(\frac{1}{2} - x\right) + \frac{1}{2}
    \]
    satisfies:
    \[
    P(x) + P(1-x) = Q\left(\frac{1}{2} - x\right) + \frac{1}{2} + Q\left(\frac{1}{2} - (1-x)\right) + \frac{1}{2} = Q\left(\frac{1}{2} - x\right) + Q\left(x - \frac{1}{2}\right) + 1
    \]
    Since \( Q \) is odd, \( Q\left(x - \frac{1}{2}\right) = -Q\left(\frac{1}{2} - x\right) \), thus:
    \[
    P(x) + P(1-x) = Q\left(\frac{1}{2} - x\right) - Q\left(\frac{1}{2} - x\right) + 1 = 1
    \]
    which confirms that \( P(x) \) satisfies the functional equation.

The final answer is \( \boxed{ P(x) = Q\left(\frac{1}{2} - x\right) + \frac{1}{2} } \) where \( Q \) is an odd polynomial of degree 2015.