Let denote $A=P(a),\, B=P(b),\, \delta=a-b$. We have $a-b | P(a)-P(b)$, i.e. (1) $A-B = k\delta ,\, k \in \mathbb{Z} $. (2) $ A.B = -\delta^2$. Eliminating $A$ from (1) and pluging it in (2) we get: $B^2+k\delta B + \delta^2 =0$. It follows that $ D=(k^2-4)\delta^2$ is a perfect square, which could be only when $k=\pm2$. In this case we obtain: $ (A+B)^2= (A - B)^2 + 4AB= 4\delta^2 - 4\delta^2= 0 \,\Rightarrow \, A+B=0 $

Already been posted here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=38&t=424938

Can also be done as follows: Let $P(a)=d$ and $P(b)=-\frac{(b-a)^2}{d}$. Then, by Bezout, we have $b-a\mid P(b)-P(a)$ implying that, $b-a\mid \frac{(b-a)^2}{d} +d$. Therefore, the object, $\frac{b-a}{d}+\frac{d}{b-a}\in\mathbb{Z}$. Now, I claim that, if $x,y$ are integers with $\varphi(x,y)=\frac{x}{y}+\frac{y}{x}\in\mathbb{Z}$, then $|x|=|y|$. To see this, simply let $g={\rm gcd}(x,y)$ with $x=gx_1,y=gy_1$. This yields, $\varphi(x,y)=\varphi(x_1,y_1) = \frac{x_1^2+y_1^2}{x_1 y_1}$. Thus, $y_1\mid x_1^2$ and $x_1\mid y_1^2$, implying $|x_1|=|y_1|=1$. Applying this observation, we get $|d|=|b-a|$, and thus, $d=b-a$ or $d=-(b-a)$, in both cases, we have $P(a)+P(b)=0$. Edit. To add, the original source of this problem is apparently a MM magazine problem, proposed by Bjorn Poonen of MIT.