Peter wrote: Suppose $p(x) \in \mathbb{Z}[x]$ and $P(a)P(b)=-(a-b)^{2}$ for some distinct $a, b \in \mathbb{Z}$. Prove that $P(a)+P(b)=0$. $P(x)=a_{n}x^{n}+...+a_{1}x+a_{0}$ => \[P(a)-P(b)=a_{n}(a^{n}-b^{n})+...+a_{1}(a-b)\] => \[(a-b)|(P(a)-P(b))\] => \[P(a)P(b)|(P(a)-P(b))^{2}\] (we suppose that $P(a)P(b)$ is not equal $0$,otherwise we would have that $P(a)=P(b)=0$)=> \[P(a)P(b)|(P(a)^{2}+P(b)^{2})\] => \[P(a)|P(b)\] \[P(b)|P(a)\] => \[|P(a)|=|P(b)|\] The case $P(a)=P(b)$ again gives us that $P(a)=P(b)=0$,the other case is $P(a)=-P(b)$ <=> $P(a)+P(b)=0$ :yes: .

Tiks wrote: we suppose that $P(a)P(b)$ is not equal $0$ In fact, we are even given that (since $a,b$ are distinct).

Peter wrote: Tiks wrote: we suppose that $P(a)P(b)$ is not equal $0$ In fact, we are even given that (since $a,b$ are distinct). Ooops,I forgot that .

$P(a)-P(b)$ is divided $a-b$ the rest is easy.