4
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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 $n\ge0$

$\endgroup$
1
$\begingroup$

It was not clear to me how the given four relations of the roots gave that $a=b=1$, so here is a different approach.


We have that $$ Q(x)=x^3+bx+a=0\tag{1} $$ and $$ P\left(\frac1x\right)=\frac1{x^3}+a\frac1{x^2}+b=0\tag{2} $$ have the same roots. Multiplying $(2)$ by $\dfrac{x^3}{b}$ yields $$ R(x)=x^3+\frac abx+\frac1b=0\tag{3} $$ Since $Q$ and $R$ have the same roots, so must their difference, $$ L(x)=\left(a-\frac ab\right)x+\left(a-\frac1b\right)=0\tag{4} $$ If $a-\frac ab\ne0$, $L$ has one root, and so must $Q$. If $Q$ has only one root, then because the coefficient of $x^2$ is $0$, that root must be $0$. However, since the roots of $P$ and $Q$ are reciprocals, $0$ can't be a root of either. Contradiction.

Therefore, $a-\frac ab=0$ and $a-\frac1b=0$, which means $a=b=1$.


Thus, $$ \begin{align} P(x)&=x^3+x^2+1\\ Q(x)&=x^3+x+1 \end{align}\tag{5} $$ Furthermore, $$ \gcd(P(x),Q(x))\mid (2x^2-x+1)P(x)-(2x^2+x-2)Q(x)=3\tag{6} $$ So, considering $\text{mod }3$:

If $x\equiv0\pmod{3}$, then $P(x)\equiv1\pmod{3}$
If $x\equiv1\pmod{3}$, then $P(x)\equiv Q(x)\equiv0\pmod{3}$
If $x\equiv2\pmod{3}$, then $P(x)\equiv1\pmod{3}$

Therefore, $$ \gcd(P(x),Q(x))=\left\{\begin{array}{l} 3&\text{if }x\equiv1\pmod{3}\\ 1&\text{if }x\not\equiv1\pmod{3} \end{array}\right.\tag{7} $$ Since $2013!+1\equiv1\pmod{3}$, $$ \gcd(P(2013!+1),Q(2013!+1))=3\tag{8} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.