polynomials - A question on greatest common divisor

I had this question in the Maths Olympiad today. I couldn't solve the part of the greatest common divisor. Please help me understand how to solve it. The question was this:

Let $P(x)=x^3+ax^2+b$ and $Q(x)=x^3+bx+a$, where $a$ and $b$ are non-zero real numbers. If the roots of $P(x)=0$ are the reciprocals of the roots of $Q(x)=0$, then prove that $a$ and $b$ are integers. Also find the greatest common divisor of $P(2013!+1)$ and $Q(2013!+1)$.

Let the roots of $P(x)=0$ be $\alpha,\; \beta, \;and\;\gamma$. Then we have the following four relations. $$\alpha+\beta+\gamma=-a$$ $$\alpha\beta\gamma=-b$$ $$\frac{1}{\alpha\beta}+\frac{1}{\beta\gamma}+\frac{1}{\alpha\gamma}=b$$ $$\frac{1}{\alpha\beta\gamma}=-a$$

From these, we get $a=b=1$ So, $P(x)=x^3+x^2+1$ and $Q(x)=x^3+x+1$. Now, how to proceed further?

Answer

If $d$ divides $P(x),Q(x)$

$d$ will divide $P(x)-Q(x)=x^2-x=x(x-1)$

But $d$ can not divide $x$ as $(P(x),x)=1\implies d$ will divide $x-1$

Again, $d$ will divide $x^2+x+1-(x^2-x)=2x+1$

Again, $d$ will divide $2x+1-2(x-1)=3$

Observe that $3$ divides $P(2013!+1),Q(2013!+1)$

as $2013!+1\equiv1\pmod3, (2013!+1)^n\equiv1$ for any integer $n\ge0$