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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.

MrAP
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2 Answers2

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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 $|x||y|=0$.

Ghartal
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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 $\;\cdots$

dxiv
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  • The quadratic equation in $a$ has two complex roots since the discriminant is negative. We can similarly rearrange the same equation and regard it as a quadratic equation in $b$ and from it we obtain two complex values of b and thus there are two solution pairs $(a,b)$ possible. Right? – MrAP Jun 01 '17 at 19:00
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    @MrAP The question asks for real roots. The only case where $\Delta$ is not negative is for $b=1$ when $\Delta=0$, therefore $b=1$, and in that case the equation reduces to $a^2-2a+1=0$. – dxiv Jun 01 '17 at 19:07