Problem 1576

1. **Fix $y$ and consider $h(x, y) = \gcd(f(x), f(y))$:**
- Since $h(x, y)$ is defined as the greatest common divisor of $f(x)$ and $f(y)$, it must be a positive integer that divides $f(y)$.
- Therefore, $h(x, y) \leq f(y)$ for all $x$.

2. Bounded polynomial argument:
- Since $h(x, y) \leq f(y)$ for all $x$, $h(x, y)$ is bounded above by $f(y)$.
- A polynomial in $x$ that is bounded for all $x$ must be constant. This is because a non-constant polynomial in $x$ would grow without bound as $x$ increases or decreases.

3. **Conclusion for fixed $y$:**
- Since $h(x, y)$ is a polynomial in $x$ and is bounded, it must be constant for fixed $y$.
- Let this constant be $c_y$. Thus, $h(x, y) = c_y$ for all $x$.

4. **Symmetry and polynomial in $y$:**
- By the definition of $h$, we have $h(x, y) = h(y, x)$. Therefore, $c_y = h(x, y) = h(y, x) = c_x$.
- This implies that $c_y$ does not actually depend on $y$ and is a constant value $c$ for all $x$ and $y$.

5. Final conclusion:
- Since $h(x, y)$ is constant for all $x$ and $y$, it follows that $h(x, y)$ is a constant polynomial.

$\blacksquare$