Mathematical Reflections 2012, Issue 6 - Problem J247

Problem:
Let $a$ and $b$ be distinct zeros of the polynomial $x^3-2x+c$. Prove that $a^2(2a^2+4ab+3b^2)=3$ if and only if $b^2(3a^2+4ab+2b^2)=5$.

Proposed by Titu Andreescu.

Solution:

Let $\alpha$ be the other root of the polynomial $x^3-2x+c$. Then $\alpha^3-2\alpha+c=0$ and $$a+b=-\alpha, \qquad ab=\alpha^2-2.$$ Therefore, $$\begin{array}{rcl} a^2(2a^2+4ab+3b^2)+b^2(3a^2+4ab+2b^2) & = & 2(a+b)^2(a^2+b^2)+2a^2b^2 \\ & = & 2\alpha^2 [\alpha^2-2(\alpha^2-2)]+2(\alpha^2-2)^2 \\ &=& 2[\alpha^2-(\alpha^2-2)]^2 \\ &=& 8, \end{array}$$ and the statement follows.