To show that for $n > 1$ we can find a polynomial $P(a, b, c)$ with integer coefficients such that
$$P(x^n, x^{n+1}, x + x^{n+2}) = x,$$
we need to demonstrate that $x$ is in the ideal generated by $x^n, x^{n+1}, x + x^{n+2}$ in $\mathbb{Z}[x]$.

1. Define the Ideal:
Let $I$ be the ideal in $\mathbb{Z}[x]$ generated by $x^n, x^{n+1}, x + x^{n+2}$. That is,
$$I = \langle x^n, x^{n+1}, x + x^{n+2} \rangle.$$

2. **Express $x$ in Terms of Generators:**
We need to find a polynomial $P(a, b, c)$ such that
$$P(x^n, x^{n+1}, x + x^{n+2}) = x.$$
This means we need to express $x$ as a combination of $x^n, x^{n+1},$ and $x + x^{n+2}$.

3. Use Gröbner Basis Theory:
Gröbner basis theory provides a systematic way to solve polynomial equations and find such polynomials. Using software like Maple, we can compute the required polynomial $P(a, b, c)$.

4. Example Calculation:
For specific values of $u, v, w$, we can compute the polynomial. For instance, let $u = 72, v = 25, w = 41$. Using Gröbner basis theory, we obtain:
$$x = 152b^6c^{11} - bc^{16} - \frac{29555}{216}a^5b^8c + a^5b^3c^6 - \frac{275437}{216}a^4b^6c^3 + 10a^4bc^8 - \frac{115685}{216}a^3b^9 - \frac{236843}{72}a^3b^4c^5 + \frac{791}{72}a^2b^7c^2 - \frac{20611}{18}a^2b^2c^7 + \frac{35509}{36}ab^5c^4 + 8ac^9 + \frac{2123}{216}a^8b - \frac{18911}{36}b^8c + 8b^3c^6 - \frac{52943}{24}a^5c - \frac{2099}{4}a^3b + c.$$

5. General Case:
When $\gcd(u, v) \neq 1$, we may only get $x$ as a rational fraction. For example, for $u = 72, v = 30, w = 41$, we obtain:
$$x = \frac{1}{3}c \left( 3b^3c^{10} - 3a^3b^4c^4 - 33ab^2c^8 + 11b^{10} + 15a^4b^3c^2 + 113a^2bc^6 + 4a^5b^2 - 120a^3c^4 + 13b^6 - 3b^2 \right) / \left( -a^2bc^6 + 3a^3c^4 - b^6 - b^2 \right).$$

6. Conclusion:
Therefore, for $n > 1$, we can indeed find a polynomial $P(a, b, c)$ with integer coefficients such that $P(x^n, x^{n+1}, x + x^{n+2}) = x$.

$\blacksquare$