Problem 1179

To determine if there is a non-natural natural number $n$ such that $\sqrt{n + 1} + \sqrt{n - 1}$ is rational, we can proceed as follows:

1. Let $\sqrt{n + 1} + \sqrt{n - 1} = \frac{a}{b}$, where $\frac{a}{b}$ is a rational number in its simplest form (i.e., $a$ and $b$ are coprime integers).

2. Square both sides of the equation:
$$\left( \sqrt{n + 1} + \sqrt{n - 1} \right)^2 = \left( \frac{a}{b} \right)^2$$
This gives:
$$n + 1 + n - 1 + 2\sqrt{(n + 1)(n - 1)} = \frac{a^2}{b^2}$$
Simplifying, we get:
$$2n + 2\sqrt{(n + 1)(n - 1)} = \frac{a^2}{b^2}$$

3. Isolate the square root term:
$$2\sqrt{(n + 1)(n - 1)} = \frac{a^2}{b^2} - 2n$$
$$\sqrt{(n + 1)(n - 1)} = \frac{\frac{a^2}{b^2} - 2n}{2}$$

4. For $\sqrt{(n + 1)(n - 1)}$ to be rational, the expression $(n + 1)(n - 1)$ must be a perfect square. Let $(n + 1)(n - 1) = k^2$ for some integer $k$. This simplifies to:
$$n^2 - 1 = k^2$$
$$n^2 - k^2 = 1$$
$$(n - k)(n + k) = 1$$

5. The equation $(n - k)(n + k) = 1$ has only two integer solutions: $n - k = 1$ and $n + k = 1$. Solving these, we get:
$$n - k = 1 \quad \text{and} \quad n + k = 1$$
Adding these two equations, we get:
$$2n = 2 \implies n = 1$$
Subtracting these two equations, we get:
$$2k = 0 \implies k = 0$$

6. However, $n = 1$ is not a non-natural natural number (it is a natural number). Therefore, there is no non-natural natural number $n$ that satisfies the given condition.