Cute polynomial problem ! Part a):The answer is yes ! Example: Let $P(X)=aX^{1012}+b$ and $Q(X)=bX^{1012}+a$ and we need that $ab=1$ and $a^2+b^2=0$ or $a^4+1=0$ so take for example $a=cos\frac{\pi}{4}+isin \frac{\pi}{4}$ and $b=\frac{1}{a}$ and it works Part b): take $\omega=cos \frac{2\pi}{3}+isin \frac{2\pi}{3}$ Denote $a=a_0+a_3+\dots+ a_{1011}$ $b=a_1+a_4+\dots +a_{1012}$ $c=a_2+a_5+\dots+ a_{1010}$ Plugging $X=\omega$ we have : $$(a+b \omega+c \omega^2)(b+a \omega+c \omega^2)=-\omega$$and denote $x=a-c$ and $y=b-c$ when we open the paranthesis we have : $x^2+y^2-xy=-1$ which is obviously a contradiction($x^2+y^2-xy\ge 0$)