## Monday, October 26, 2015

### Heuristic Solution: Determine the product of $q(r_1)\cdot q(r_2)\cdot q(r_3)\cdot q(r_4)\cdot q(r_5)$.

Given $p(x)=x^5+x^2+1$ have roots $r_1,\,r_2,\,r_3,\,r_4,\,r_5$. Let $q(x)=x^2-2$. Determine the product of $q(r_1)\cdot q(r_2)\cdot q(r_3)\cdot q(r_4)\cdot q(r_5)$.

This is one of the greatest IMO problem that only the intelligent and who can see the subtext of everything can solve it beautifully, and that solver is my dearest math friend from the U.K., a retired Professor of Mathematics from the U.K. He is someone who is always so willing to teach me the things that left me puzzled, and therefore I've learned a great deal from him. I will always love him for being such a sweet and nice and kind and smart mentor of mine! :D

If $y=x^2-2$, then $x=(y+2)^{\frac{1}{2}}$, and

\begin{align*}x^5+x^2+1&=((y+2)^{\frac{1}{2}})^5+((y+2)^{\frac{1}{2}})^2+1\\&=(y+2)^{\frac{5}{2}}+y+2+1\\&=(y+2)^{\frac{5}{2}}+y+3\end{align*}

But if $(y+2)^{\frac{5}{2}}+y+3=0$ then we have $(y+2)^5=(y+3)^2$, so that $y^5+\cdots+(2^5-3^2)=0$.

Therefore, the product of the roots of the polynomial $q(x)$ is the negative of the constant term, i.e. $-(2^5-3^2)=-23$.