Problem 1364

1. **Define the pair $(x, y)$:**
We start by choosing $x$ such that $x \ge 2$. We then define $y$ as:
$$y = x^{n+1} - x^n - 1$$
This ensures that $1 < x < y$ for any integer $n > 1$.

2. **Calculate $y + x^n$:**
$$y + x^n = (x^{n+1} - x^n - 1) + x^n = x^{n+1} - 1$$

3. **Calculate $x + y^n$:**
$$x + y^n = x + (x^{n+1} - x^n - 1)^n$$
We need to show that $x^{n+1} - 1$ divides $x + (x^{n+1} - x^n - 1)^n$.

4. **Simplify $(x^{n+1} - x^n - 1)^n$:**
Let $z = x^{n+1} - x^n - 1$. Then:
$$x + z^n = x + (x^{n+1} - x^n - 1)^n$$
We need to show that $x^{n+1} - 1$ divides $x + z^n$.

5. **Expand $z^n$:**
$$z^n = (x^{n+1} - x^n - 1)^n$$
Using the binomial theorem, we expand $z^n$:
$$z^n = \sum_{k=0}^{n} \binom{n}{k} (x^{n+1} - x^n)^{n-k} (-1)^k$$
Notice that each term in the expansion of $z^n$ is a multiple of $x^{n+1} - x^n$ except for the constant term.

6. Simplify the expression:
$$x + z^n = x + \sum_{k=0}^{n} \binom{n}{k} (x^{n+1} - x^n)^{n-k} (-1)^k$$
Since $x^{n+1} - x^n$ is a factor in each term of the sum except for the constant term, we can write:
$$x + z^n = x + (x^{n+1} - x^n - 1)^n = x + (x^{n+1} - 1 - x^n)^n$$

7. Show divisibility:
We need to show that $x^{n+1} - 1$ divides $x + (x^{n+1} - 1 - x^n)^n$. Notice that:
$$x + (x^{n+1} - 1 - x^n)^n = x + (x^{n+1} - 1 - x^n)^n$$
Since $x^{n+1} - 1$ is a factor of $x^{n+1} - 1 - x^n$, it follows that $x^{n+1} - 1$ divides $x + (x^{n+1} - 1 - x^n)^n$.

8. Conclusion:
Therefore, for any integer $n > 1$, there exist infinitely many pairs $(x, y)$ of integers $1 < x < y$ such that $x^n + y \mid x + y^n$.

$\blacksquare$