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Find all values of $a,b$ for which the system of equations $xyz+z=a,\quad xyz^2+z=b,\quad x^2+y^2+z^2=4$ has only one real solution.


$xyz+z=a$
$xyz^2+z=b$
So,$\frac{xy+1}{xyz+1}=\frac{a}{b}$
I can think no method by so as this system of equations has only one real solution. What should I do? Please help me.

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If $(x, y, z)$ is a solution, then $(y, x, z)$ is a solution. Hence if there is a unique solution, we need $x=y$. This reduces the problem to $2$ variables.

$$\begin{align*} x^2z+z&=a\\ x^2z^2+z&=b\\ 2x^2+z^2&=4\\ \end{align*}$$

If $(x, x, z)$ is a solution, note that $(-x, -x, z)$ is also a solution. Hence, $x=0$ if the solution is unique.

$$\begin{align*} z&=a\\ z&=b\\ z^2&=4\\ \end{align*}$$

Hence $a=b$. For there to be a solution, we need $a=b=2$ or $a=b=-2$.

We now check the $2$ solutions of $(a, b)$ by solving the equations.

$$\begin{align*} xyz+z&=a\\ xyz^2+z&=b\\ x^2+y^2+z^2&=4\\ \end{align*}$$

Since $a=b$, we subtract the first equation from the second equation to get $xy(z^2-z)=0$. There are a few cases:

If $z=0$, then $xyz+z=0\neq a$.

Otherwise, if $x=0$ (or $y=0$ symmetrically), we have $z=a$ and $z^2=4$, which, substituting into the third equation, means that $y^2=0$ (or $x^2=0$), which means that $(x, y, z)=(0, 0, a)$ is one solution.

The final case $z=1$ should have no solutions. Let's check the $2$ cases $a=b=2$ and $a=b=-2$.

If $a=b=2$, the system of equations reduces to:

$$\begin{align*} xy&=1\\ x^2+y^2&=3\\ \end{align*}$$

Solving, we find another solution, $(x, y)=\left(\frac{\sqrt{5}-1}{2}, \frac{\sqrt{5}+1}{2}\right)$, so $a=b=2$ does not admit a unique solution.

If $a=b=-2$, the system of equations reduces to:

$$\begin{align*} xy&=-3\\ x^2+y^2&=3\\ \end{align*}$$

Note that $(x+y)^2=x^2+y^2+2xy=-3$. As such $x+y$ is not real and there is no extra solution here.

Hence the only $(a, b)$ for the system to have a unique solution is $(-2, -2)$.

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  • $\begingroup$ @diya See edit of answer. $\endgroup$ – Element118 Dec 30 '15 at 3:17
  • $\begingroup$ Thank you very much.I got it.@Element118 $\endgroup$ – diya Dec 30 '15 at 3:23
  • $\begingroup$ I posted an answer which is essentially a comment on your posting, but was too long to place here. @Element118 $\endgroup$ – zyx Dec 30 '15 at 4:49
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The symmetries $(x,y,z) \to (y,x,z)$ and $(x,y,z) \to (-x,-y,z)$ generate

  • $4$ solutions from $1$ if $x \neq y$
  • $2$ solutions from $1$ if $x = y \neq 0$
  • no new solutions if $x=y=0$.

Thus, the only hope for a unique solution is when $x=y=0$ which simplifies equations to $z=a=b$ and $z^2 = 4$.

This is a calculation-free reorganization of Element118's answer and the rest is the same as there, checking the cases $a=b=\pm 2$ to see which has a unique solution.

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