How to find the number of solution pairs for this equation? Question: Consider the equation $(1+a+b)^2=3(1+a^2+b^2)$ where $a$ and $b$ are real numbers. How many solution pair(s), ($a$,$b$), are possible for this equation?
My attempt: 
$(1+a+b)^2=3(1+a^2+b^2)$
=>$1+a^2+b^2+2(a+ab+b)=3+3a^2+3b^2$
=>$1+a^2+b^2+2a+2ab+2b=3+3a^2+3b^2$
=>$2a+2ab+2b=2+2a^2+2b^2$
=>$2(a+ab+b)=2(1+a^2+b^2)$
=>$a+ab+b=1+a^2+b^2$
=>$a-ab+b-1=a^2+b^2-2ab$
=>$-a(b-1)+1(b-1)=(a-b)^2$
=>$(b-1)(1-a)=(a-b)^2$
I have simplified the given equation to this point. I cannot understand how to proceed. I would be grateful for any help i receive.

Hint:   after expanding and collecting, the equation can be written as:
$$a^2 - a b - a + b^2 - b + 1 = 0 \;\;\iff\;\; a^2 - (b+1)\,a + b^2-b+1=0$$
Considering it as a quadratic in $\,a\,$, its discriminant is:
$$
\Delta=(b+1)^2-4(b^2-b+1)=-3b^2+6b-3=-3(b-1)^2
$$
For the quadratic to have real roots the discriminant must be non-negative, so $\;\cdots$
