How many real solution pairs $(a,b)$ are there for the equation $(1+a+b)^2=3(1+a^2+b^2)$?

Question

How many real solution pairs $(a,b)$ are there for the equation $(1+a+b)^2=3(1+a^2+b^2)$?

On simplification, we get $a+b+ab=1+a^2+b^2$. How to proceed further? It seems to me(from this form)that there are infinitely many solutions.

Answer 1

Write the simplified condition as $f(a,b) = -(a+b+ab)+ 1+a^2+b^2 = 0$.

Let $a = x+1$ and $b=y+1$. Then you get $f(x,y) = 1/2 [(x-y)^2 +x^2 + y^2] =0$.

Since all square terms cannot become negative, they must all be zero. Hence you have $x=y=0$ or $a=b=1$ as the only solution.

Answer 2

Fix b we can solve the equation for a. The discriminant is $(b+1)^2-4(b^2-b+1) = b^2+2b+1-4b^2+4b-4=-3b^2+6b-3=-3(b-1)^2$ This is negative for $b\not =1$ and so there is no solution. If b=1 then we have $2a+1=a^2+2$ we get that $a=1$. And so there is only one solution $a=1,b=1$

Answer 3

Alternatively using Cauchy-Schwarz inequality: $(1+a+b)^2 = (1\cdot 1 + 1\cdot a+1\cdot b)^2 \le (1^2+1^2+1^2)(1^2+a^2+b^2) = 3(1+a^2+b^2)$, with equality occurs when $\dfrac{1}{1} = \dfrac{1}{a} = \dfrac{1}{b}\implies a = b = 1$, which is also the solution.

Answer 4

Two proofs for the price of one:

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Consider the function $f(a,b) = 3(1 + a^2 + b^2) - (1 + a + b)^2 = 2(1 - a - b - ab + a^2 + b^2)$. We wish to find all values of $a$ and $b$ for which this function vanishes. The gradient of this function is $$ \nabla f = 2(-1 - b + 2a, -1 - a + 2b), $$ which can be shown to vanish only at $a = b = 1$. Moreover, the Hessian matrix is $$ \begin{bmatrix} \frac{\partial^2 f}{\partial a^2} & \frac{\partial^2 f}{\partial a \partial b} \\ \frac{\partial^2 f}{\partial a \partial b} & \frac{\partial^2 f}{\partial b^2} \end{bmatrix} = \begin{bmatrix} 4 & -2 \\ -2 & 4\end{bmatrix}, $$ which is positive definite. Thus, $f$ has a single global minimum at $a = b = 1$. But $f(1,1) = 0$, meaning that the only solution to $f(a,b) = 0$ is $a = b = 1$.

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Define $u$ and $v$ such that $a = u + v$ and $b = u - v$. Plugging these values in, the equation becomes $$ 3 + 6 u^2 + 6 v^2 = (1 + 2u )^2 \quad \Rightarrow \quad 6 v^2 + 2 (u^2 - 2u + 1) = 0. $$ This can only be satisfied when $u = 1$ and $v = 0$, i.e., $a = b = 1$.