Mathematical Reflections 2014, Issue 1 - Problem O289

Problem:
Let $a, b, x, y$ be positive real numbers such that $x^2-x+1=a^2$, $y^2+y+1=b^2$, and
$(2x-1)(2y+1)=2ab+3$. Prove that $x+y=ab$.

Proposed by Titu Andreescu.

Solution:
Multiplying both sides of the first two equations by $4$ and both sides of the third equation by $2$ , we have
$$\begin{array}{rll} (2x-1)^2+3&=&4a^2            (1)\\ (2y+1)^2+3&=&4b^2            (2)\\ 2(2x-1)(2y+1)&=&4ab+6     (3)\\ \end{array}$$
Summing up these equations, we get $$(2x-1)^2+2(2x-1)(2y+1)+(2y+1)^2+6=4(a^2+ab+b^2)+6,$$ i.e. $$[(2x-1)+(2y+1)]^2=4(a^2+ab+b^2),$$ which gives $x+y=\sqrt{a^2+ab+b^2}$. So, it suffices to show that $\sqrt{a^2+ab+b^2}=ab$. Multiplying $(1)$ and $(2)$, we get $[(2x-1)^2+3][(2y+1)^2+3]=16a^2b^2$, and using $(3)$ we get $$[(2ab+3)^2+3(4a^2-3+4b^2-3)+9]=16a^2b^2,$$ which gives $a^2+ab+b^2=a^2b^2$. Taking the square root, we obtain the conclusion.