Question

Let $P\left(x\right)=x^3+a{x}^2+b$ and $Q\left(x\right)=x^3+bx+a$, where $a,b$ are non-zero real numbers. Suppose that the roots of the equation $P\left(x\right)=0$ are the reciprocals of the roots of the equation $Q\left(x\right)=0.$ Prove that $a$ and $b$ are integers. Find the greatest common divisor of $P(2013!+1)$ and $Q(2013!+1).$

• A. $1$
• B. $2$
• C. $3$
• D. $5$

Answer 1

Answer: (C)

$3$

Note that P(0) $\ne$ 0.
Let $R\left(x\right)=x^{3}P\left(1/x\right)=b{x}^3+ax+1$.
Then the equations Q(x) = 0 and R(x) = 0 have the same roots.
$\to R\left(x\right)=bQ\left(x\right)$ and equating the coefficients we get a = $b^2$ and $ab=1$.
This implies that $b^3=1$, so a = b = 1.
$\therefore P\left(x\right)=x^3+x^2+1$ and $Q\left(x\right)=x^3+x+1$.
For any integer n we have (P(n), Q(n)) = (P(n), P(n) - Q(n)) = $(n^3+n^2+1,n^2-n)=(n^3+n^2+1,n-1)=(3,n-1)$.
$\therefore \left(P\right(n),Q(n\left)\right)=3$, if n - 1 is divisible by 3.
Since $3$ divides $2013!$ it follows that $\left(P\right(2013!+1),Q(2013!+1\left)\right)=3.$