KVPY 2016 SA Question 20

Better solution from Md Zuhair

$\begin{aligned} (1+a+b)^2 & = 3(1+a^2+b^2) \\ 1 + a^2 + b^2 + 2a + 2ab + 2b & = 3 + 3a^2 + 3b^2 \\ 2a^2 + 2b^2 - 2ab - 2a - 2b + 2 & = 0 \\ a^2 -2ab + b^2 + a^2 -2a + 1 + b^2 - 2b + 1 & = 0 \\ (a-b)^2 + (a-1)^2 + (b-1)^2 & = 0 \end{aligned}$

We note that the LHS is $\ge 0$ and is equal to the RHS only when $a=b=1$ . Therefore, there is exactly one solution pair $(a,b)$ $\ = (1,1)$ .

My solution

$\begin{aligned} (1+a+b)^2 & = 3(1+a^2+b^2) \\ 1 + a^2 + b^2 + 2a + 2ab + 2b & = 3 + 3a^2 + 3b^2 \\ 2a^2 + 2b^2 - 2ab - 2a - 2b + 2 & = 0 \\ a^2 + b^2 - ab - a - b + 1 & = 0 \\ b^2 - (a+1)b + a^2 - a + 1 & = 0 & \small \color{#3D99F6} \text{A quadratic equation of }b \end{aligned}$

$\begin{aligned} \implies b & = \frac {(a+1) \pm \sqrt{(a+1)^2-4a^2+4a-4}}2 \\ & = \frac {(a+1) \pm \sqrt{a^2 + 2a+1-4a^2+4a-4}}2 \\ & = \frac {(a+1) \pm \sqrt{-3a^2 +6a-3}}2 \\ & = \frac {(a+1) \pm \sqrt{-3(a^2 +2a-1)}}2 \\ & = \frac {(a+1) \pm \sqrt{-3(a-1)^2}}2 \\ & = \frac {(a+1) \pm {\color{#3D99F6}\sqrt{-3}(a-1)}}2 \end{aligned}$

Note that $b$ is only real when ${\color{#3D99F6}\sqrt{-3}(a-1)} = 0$ , that is when $a=1$ and $b=1$ . Therefore, there is exactly one solution pair $(a,b)$ $\ = (1,1)$ .