They Have Something In Common!

Let $f(x)=x^2+ax+b$ and let the common root be $r$ . So $f(r)=0$ and $f(f(f(r)))=0$ . This implies $f(f(0))=0$ . But $f(0)=b$ so $f(b)=0$ . So $b$ is the other root. Since the product of the two roots is $b$ , we have $rb=b$ which implies $b=0$ or $r=1$ . These respectively implies $f(0)=0$ or $f(1)=0$ . Both leads to the whole product to be $\fbox{0}$ .

$f(x)=x^2$ was the first thing which came into my mind!

it always implies that f(0) = a, then f(2014) = 0