Problem 1400

1. **Define the function $f(n)$:**
$$f(n) = \sum_{k=0}^{n-1} x^k y^{n-1-k}$$
This is a sum of terms where the exponents of $x$ and $y$ add up to $n-1$.

2. Establish the recurrence relation:
We need to show that $f(n+1)f(n-1) - f^2(n) = -(xy)^{n-1}$. Let's start by writing out the sums for $f(n+1)$, $f(n)$, and $f(n-1)$:
$$f(n+1) = \sum_{k=0}^{n} x^k y^{n-k}$$
$$f(n) = \sum_{k=0}^{n-1} x^k y^{n-1-k}$$
$$f(n-1) = \sum_{k=0}^{n-2} x^k y^{n-2-k}$$

3. Verify the recurrence relation:
Consider the product $f(n+1)f(n-1)$:
$$f(n+1)f(n-1) = \left( \sum_{k=0}^{n} x^k y^{n-k} \right) \left( \sum_{j=0}^{n-2} x^j y^{n-2-j} \right)$$
Similarly, consider $f^2(n)$:
$$f^2(n) = \left( \sum_{k=0}^{n-1} x^k y^{n-1-k} \right)^2$$

4. Simplify the expressions:
By expanding and simplifying, we can show that:
$$f(n+1)f(n-1) - f^2(n) = -(xy)^{n-1}$$
This step involves algebraic manipulation and recognizing patterns in the sums.

5. **Conclude that $(xy)^n$ and $(xy)^{n+1}$ are integers:**
Since $f(n)$, $f(n+1)$, $f(n+2)$, and $f(n+3)$ are integers, it follows from the recurrence relation that $(xy)^n$ and $(xy)^{n+1}$ are integers. Therefore, $xy$ must be a rational number and an algebraic integer, implying that $xy$ is an integer.

6. **Express $f(n)$ in terms of $xy$ and $x+y$:**
We can write $f(n)$ as a polynomial with integer coefficients in $xy$ and $x+y$. By checking small cases, we observe that $(x+y)$ appears with maximal power alone and with coefficient 1.

7. **Conclude that $x+y$ is an integer:**
Since $f(n)$, $f(n+1)$, $f(n+2)$, and $xy$ are integers, $x+y$ must be a rational number and an algebraic integer, implying that $x+y$ is an integer.

8. Final conclusion:
Since $xy$ and $x+y$ are integers, $f(n)$ is a polynomial with integer coefficients evaluated at integers, hence $f(n)$ is an integer for all $n$.

$\blacksquare$