2024 SSMO Team Round Problems/Problem 14

Problem

Let be the roots of the polynomial Find the value of

Solution

Let denote the polynomial. We have since covers the case of (which is the domain of the RHS product) squaring the nested product doubles it, covering the symmetric case, and the third factor covers the case. Now, note that \begin{align*} \prod_{m=1}^{7}(a_na_m-1)&=-a_n^7\prod_{m=1}^7\left(\frac{1}{a_n}-a_m\right)\\ &=-a_n^7f\left(\frac{1}{a_n}\right)\\ &=-a_n^7-(5a_n^6+10a_n^5+a_n^4+a_n^3+9a_n^2+5a_n+1)\\ &=(5a_n^6+9a_n^5+a_n^4+a_n^3+10a_n^2+5a_n+1)-(5a_n^6+10a_n^5+a_n^4+a_n^3+9a_n^2+5a_n+1)\\ &=-a_n^5+a_n^2\\ &=-(a_n)^2(a_n^3-1). \end{align*} So, \begin{align*} \prod_{n=1}^7\prod_{n=1}^7(a_na_m-1)&=\prod_{n=1}^7\left(-(a_n)^2(a_n^3-1)\right)\\ &=-\left(\prod_{n=1}^7a_n\right)^2\prod_{n=1}^7(a_n-1)\prod_{n=1}(a_n-e^{\frac{2\pi i}{3}})\prod_{n=1}(a_n-e^{\frac{4\pi i}{3}})\\ &=-(-1)^2(-f(1))\left(-f\left(e^{\frac{2\pi i}{3}}\right)\right)\left(-f\left(e^{\frac{4\pi i}{3}}\right)\right)\\ &=f(1)f\left(e^{\frac{2\pi i}{3}}\right)f\left(e^{\frac{4\pi i}{3}}\right). \end{align*} For and So, \begin{align*} f\left(e^{\frac{2\pi i}{3}}\right) &= 12\left(\frac{-1+i\sqrt{3}}{2}\right)^2 = \frac{-12-12i\sqrt{3}}{2} = -6-6i\sqrt{3}\text{ and }\\ f\left(e^{\frac{4\pi i}{3}}\right) &= 12\left(\frac{-1-i\sqrt{3}}{2}\right)^2 = \frac{-12+12i\sqrt{3}}{2} = -6+6i\sqrt{3}\implies\\ f\left(e^{\frac{2\pi i}{3}}\right)f\left(e^{\frac{4\pi i}{3}}\right) &= (-6-6i\sqrt{3})(-6+6i\sqrt{3}) = 144. \end{align*} Thus, Now, we have \begin{align*} \prod_{n=1}^7(a_n^2-1)&=\prod_{n=1}^7(a_n-1)\prod_{n-1}^7(a_n+1)\\ &=(-f(1))(-f(-1))\\ &=f(1)f(-1)\\ &=33\cdot1 = 33. \end{align*} Substituting this into we have

~SMO_Team