Problem 1400

1. Let $x = (\alpha + \sqrt{\alpha^2 - 1})^{1/n}$ and $y = (\alpha - \sqrt{\alpha^2 - 1})^{1/n}$.

2. We observe that:
$$x^n = \alpha + \sqrt{\alpha^2 - 1} \quad \text{and} \quad y^n = \alpha - \sqrt{\alpha^2 - 1}$$

3. Multiplying these two expressions, we get:
$$x^n y^n = (\alpha + \sqrt{\alpha^2 - 1})(\alpha - \sqrt{\alpha^2 - 1}) = \alpha^2 - (\alpha^2 - 1) = 1$$
Therefore, $x^n y^n = 1$, which implies $(xy)^n = 1$. Since $x$ and $y$ are real numbers, $xy = \pm 1$.

4. Assume, for the sake of contradiction, that $x + y$ is rational.

5. By the binomial theorem and properties of polynomials, we can show by induction that $x^k + y^k$ can be written as a polynomial in $xy$ and $x + y$ with integer coefficients for $k \in \mathbb{N}$.

6. Specifically, for $k = n$, we have:
$$x^n + y^n = (\alpha + \sqrt{\alpha^2 - 1}) + (\alpha - \sqrt{\alpha^2 - 1}) = 2\alpha$$
Since $x + y$ is assumed to be rational, and $xy = \pm 1$, $x^n + y^n$ would also be rational.

7. However, $x^n + y^n = 2\alpha$, and since $\alpha$ is irrational, $2\alpha$ is also irrational. This is a contradiction because we assumed $x + y$ is rational, leading to $x^n + y^n$ being rational.

8. Therefore, our assumption that $x + y$ is rational must be false. Hence, $x + y$ is irrational.

$\blacksquare$