Solve the following set of equations.
$\displaystyle \begin{cases} x+y+z=a \\ x^2+y^2+z^2 = b^2 \\ xy=z^2\end{cases}$
I have tried to solve them by substitution but did not get anything useful.
$\displaystyle \begin{align*} x+y+z &=a \quad [1] \\ (x+y+z)^2 &= a^2 \\ x^2+y^2+z^2+2(xy+xz+yz)&=a^2 \\ b^2+2[z^2+z(x+y)] &= a^2 \quad [\text{from} \, 2 \, \text{and} \, 3]\\z^2+z(a-z) &=\frac{a^2-b^2}{2} \\ az &=\frac{a^2-b^2}{2} \\ z &= \frac{a^2-b^2}{2a}\end{align*}$$\displaystyle \begin{cases} x+y+z=a \quad [1]\\ x^2+y^2+z^2 = b^2 \quad [2]\\ xy=z^2 \quad[3] \end{cases}$
Similarly try finding the values of x and y.
We have got $\displaystyle z=\frac{a^2-b^2}{2a} \quad [4]$.$\displaystyle \begin{cases} x+y+z=a \quad [1] \\ x^2+y^2+z^2=b^2 \quad [2] \\ xy=z^2 \quad [3]\end{cases}$
$\displaystyle xy=z^2 \quad [3] \\ \implies y= \frac{1}{x} \left( \frac{a^2-b^2}{2a}\right)^2 \quad [5]$
$\displaystyle \begin{align*} x+y+z &=a \quad [1] \\ x+ \frac{1}{x} \left( \frac{a^2-b^2}{2a}\right)^2+ \frac{a^2-b^2}{2a} &= a \quad [\text{from} \, 4 \, \text{and} \, 5] \\x^2 -2ax(a^2+b^2)+(a^2-b^2)^2 &=0 \end{align*}$
You have a quadratic equation in x .