$P(x)=x(x-a_1)...(x-a_n)$ where $a_1,...,a_n,n \geq 4$ are different nonzero integers. $P(P(x))=P(x)(P(x)-a_1)...(P(x)-a_n)$. Obvious, that $0,a_1,...,a_n$ are roots of $P(P(x))$ Let there is another root $b$. Then for some $i$ it is true, that $P(b)=a_i$ $b(b-a_1)...(b-a_i)...(b-a_n)=a_i$ so $b(b-a_i)|a_i $ $a_i=bu \neq 0\to b(b-bu)|bu \to b(1-u)|u \to 1-u|1 \to u=2,a_i=2b$ so $(b-a_1)...(-b)...(b-a_n)=2$ but $2$ has only three different divisors, so there are not integer solutions. So answer : all roots of $P(P(x))=0$ are roots of $P(x)=0$

How are ensuring that b is integer?

amrit1968 wrote: How are ensuring that b is integer? question wants to find all integer roots of PoP so b is integer