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.

Although, dxiv showed you a standard solution, but you have a good result there. Let $x=1-a$ and $y=b-1$ then $xy=(x+y)^2$ or $-xy=x^2+y^2$. Notice that $$2|x||y| \le x^2+y^2=-xy \le |x||y|,$$which gives you $|x||y| \le 0$. This is possible only when $|x||y|=0$.
