Problem 1415

To prove that $P(x)$ is a constant polynomial, we will use the given condition that if $P(Q(x)) = F(x) \cdot G(x)$ for polynomials $F(x)$, $G(x)$, and $Q(x)$ with integer coefficients, then either $F(x)$ or $G(x)$ must be a constant polynomial.

1. **Assume $P(x)$ is not a constant polynomial**:
Let $P(x)$ be a polynomial of degree $n \geq 1$.

2. **Consider the degree of $P(Q(x))$**:
If $Q(x)$ is a polynomial of degree $m$, then $P(Q(x))$ will be a polynomial of degree $n \cdot m$.

3. **Factorization of $P(Q(x))$**:
Since $P(Q(x)) = F(x) \cdot G(x)$, the degrees of $F(x)$ and $G(x)$ must multiply to $n \cdot m$. Let the degrees of $F(x)$ and $G(x)$ be $d_1$ and $d_2$ respectively. Thus, $d_1 \cdot d_2 = n \cdot m$.

4. Given condition:
By the problem's condition, either $F(x)$ or $G(x)$ must be a constant polynomial. Without loss of generality, assume $F(x)$ is a constant polynomial. Therefore, $d_1 = 0$.

5. Implication for degrees:
Since $d_1 = 0$, we have $0 \cdot d_2 = n \cdot m$. This implies $n \cdot m = 0$.

6. **Conclusion about $n$**:
Since $m$ is the degree of $Q(x)$ and can be any non-negative integer, the only way $n \cdot m = 0$ holds for all $m$ is if $n = 0$.

7. Final conclusion:
If $n = 0$, then $P(x)$ must be a constant polynomial.

Thus, we have shown that $P(x)$ must be a constant polynomial.

$\blacksquare$