Let polynomial is $P(x)$ and $n \geq 3$ $P(x)$ has integer root $t$ and $p$ is prime divisor of $|t|$ $ t|a_n \to p=p_i$ for some $i$ and $p_i^2\not |t$ $P(t)=0 \to p_i^2|a_n+a_{n-1}t=a_{n-1}(p_n+t)\to p_i^2|p_n+t \to p_i=p_n$ Let $t=ap_n$ Then $ p_n^3| a_{n-2}t^2+a_{n-1}t+a_n=a_{n-2}( t^2+p_{n-1}t+p_{n-1}p_n) \to p_n^2|a^2p_n+ap_{n-1}+p_{n-1} \to p_n|a+1 \to p_n^2|p_n \to$ contradiction So $n<3$ For $n=1$ we have $P(x)=x+p_1$ has root $-p_1$ For $n=2$ we have $P(x)=x^2+p_1x+p_1p_2$ don`t have roots

Assume that $n\ge 3$ , and it has an integer root, so it's reducible in $\mathbb{Z[X]}$ , so it's also reducible in $\mathbb{Q[X]}$ , but note that the prime $p_1 | a_1 , a_2,.... a_n$ and $p_1 ^2$ doesn't divides $a_n$ , this means by Eisentein irreduciblity Criterion, the polynomial is irreducible in $\mathbb{Q[X]}$ , a Contradiction... Hence $n < 3$.

Why you assume n is bigger than 3, just assume n is bigger than 2 is OK . (Eisentein irreduciblity Criterion also can use during n=2 , but sometimes you must make a substitution) then n can only be 1.

We claim that $n=1$ is the only solution. This is clearly true because $x+a_1$ has root $-a_1$. For $n\geq 2$, by applying Eisenstein's Criterion on $p_1$, we know that the polynomial is irreducible over $\mathbb{Q}[X]$ and by extension $\mathbb{Z}[X]$.

If n>1, than p(x) irreducible polynomial. Answer n=1.

By Eisenstein's criterion our polynomial is irreducible over $Z$ so $n = 1$ as claimed

Eisenstein's Criterion $$answer: \boxed {n=1}$$

Let $r$ be an integer root of the polynomial, then $$r^n+p_1r^{n-1}+p_1p_2r^{n-2}+\cdots +p_1p_2\cdots p_n=0.$$Let us observe that from second term of the polynomial, the prime factor $p_1$ appears, then $r$ is a multiple of $p_1$. If $n\geq 2$, then the first $n$ term are a multiples of $p_1^2$, then $p_1^2\mid p_1p_2\cdots p_n$, which is absurd. Therefore, $n=1$ and the polynomial is $P(x)=x+p_1$, where clearly has an integer raiz, which is $-p_1$. Finally, the only value of $n$ is $1$.