An algebra problem by Prithwish Roy

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

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 $(a,b)$ $\ = (1,1)$ .

By C - S Inequality on the sets $(a , b , 1) , (1 , 1 , 1)$ , we have

$(a^2 + b^2 + 1)(1 + 1 + 1) \geq (a + b + 1)^2$

Equality holds when $a = b = 1$ .

Therefore our only solution is $\boxed{(a = b = 1)}$

Let's expand the above equation out to obtain:

$(a+b)^2 + 2(a+b) + 1 = 3 + 3a^2 + 3b^2;$

or $a^2 + b^2 + 2ab + 2a + 2b + 1 = 3 + 3a^2 + 3b^2$ ;

or $0 = 2a^2 + 2b^2 - 2a - 2b - 2ab + 2;$

or $0 = a^2 - (b+1)a + (b^2 - b + 1)$ ;

The corresponding roots are:

$a = \frac{(b+1) \pm \sqrt{(b+1)^2 - 4(1)(b^2 - b + 1)}}{2} = \frac{(b+1) \pm \sqrt{-3b^2 + 6b -3}}{2} = \frac{(b+1) \pm \sqrt{-3(b-1)^2}}{2} = \frac{(b+1) \pm (b-1)\sqrt{-3}}{2}$

which $a \in \mathbb{R}$ holds true iff $b = 1 \Rightarrow a = 1$ . Hence $\boxed{(a,b) = (1, 1)}$ is the only real pair allowed.

Apply cauchy-schwarz inequality to the LHS.We will observe that RHS obtained will be the RHS given in the question.Now apply equality condition of cauchy-schwarz inequality to get a=b=1.

If we replace $a$ by $x$ and $b$ by $y$ , and expanding this, we have $x^2+y^2-xy-x-y+1=0$ This is an equation for an ellipse. Also this satisfies the determinant which means that it is a point ellipse. Thus there is only 1 solution.

Relevant wiki: Factorization of Polynomials

It is very simple,

If we simplify the equation we get

$2a^2 + 2b^2 +2ab+ 2 - 2a - 2b = 0$

So $(a-b)^2+(a-1)^2+(b+1)^2 = 0$

Squares are zero and a and b are real numbers

Hence a=b, a+1 and b=1

So $\boxed{a=b=1} \text{is the only solution}$