
6. Let $a, b$ be real numbers and, let $P(x)=x^{3}+a x^{2}+b$ and $Q(x)=x^{3}+b x+a$. Suppose that the roots of the equation $P(x)=0$ are the reciprocals of the roots of the equation $Q(x)=0$. Find the greatest common divisor of $P(2013!+1)$ and $Q(2013!+1)$.


Solution. Note that $P(0) \neq 0$. Let $R(x)=x^{3} P(1 / x)=b x^{3}+a x+1$. Then the equations $Q(x)=0$ and $R(x)=0$ have the same roots. This implies that $R(x)=b Q(x)$ and equating the coefficients we get $a=b^{2}$ and $a b=1$. This implies that $b^{3}=1$, so $a=b=1$. Thus $P(x)=x^{3}+x^{2}+1$ and $Q(x)=x^{3}+x+1$. For any integer $n$ we have

$(P(n), Q(n))=(P(n), P(n)-Q(n))=\left(n^{3}+n^{2}+1, n^{2}-n\right)=\left(n^{3}+n^{2}+1, n-1\right)=(3, n-1)$.

Thus $(P(n), Q(n))=3$ if $n-1$ is divisible by 3 . In particular, since 3 divides 2013 ! it follows that $(P(2013!+1), Q(2013!+1))=3$.

