Let $f(x)=x^{n}+a_{n-1} x^{n-1}+\cdots+a_{0}$ be a polynomial of degree $n \geq 1$ with $n$ (not necessarily distinct) integer roots. Assume that there exist distinct primes $p_{0}, p_{1}, \ldots, p_{n-1}$ such that $a_{i}>1$ is a power of $p_{i}$, for all $i=0, \ldots, n-1$. Find all possible values of $n$.
Obviously all the roots have to be negative by the positivity of the coefficients. If at least two of the roots are unequal to -1, then both of them have to be powers of \( p_{0} \). Now Vieta's formulæ yield \( p_{0} \mid a_{1} \), which is a contradiction. Thus we can factor \( f \) as

\[
f(x) = \left(x + a_{0}\right)(x + 1)^{n-1}
\]

Expanding yields

\[
a_{2} = \left(\begin{array}{c}
n-1 \\
1
\end{array}\right) + a_{0} \left(\begin{array}{c}
n-1 \\
2
\end{array}\right) \quad \text{and} \quad a_{n-2} = a_{0} \left(\begin{array}{c}
n-1 \\
n-2
\end{array}\right) + \left(\begin{array}{c}
n-1 \\
n-3
\end{array}\right)
\]

If \( n \geq 5 \), we see that \( 2 \neq n-2 \) and so the two coefficients above are relatively prime, being powers of two distinct primes. However, depending on the parity of \( n \), we have that \( a_{2} \) and \( a_{n-2} \) are both divisible by \( n-1 \) or \( \frac{n-1}{2} \), which is a contradiction.

For \( n = 1, 2, 3, 4 \), the following polynomials meet the requirements:

\[
\begin{aligned}
& f_{1}(x) = x + 2 \\
& f_{2}(x) = (x + 2)(x + 1) = x^{2} + 3 x + 2 \\
& f_{3}(x) = (x + 3)(x + 1)^{2} = x^{3} + 5 x^{2} + 7 x + 3 \\
& f_{4}(x) = (x + 2)(x + 1)^{3} = x^{4} + 5 x^{3} + 9 x^{2} + 7 x + 2
\end{aligned}
\]