Solving algebraic equations inolving square terms

Question

How to find the values of $x,y$ and $z$ if $3x²-3(1126)x=96y²+24(124)y=8z²-4(734)z $?

I dont have any idea!! I think we can have many values of $x,y$ and $z$ at a time or it is a no solution??

Answer 1

Perhaps one way to do it is to solve the system \begin{align*} x^2 - 1126 x & = 32 y^2 + 992 y, & (1)\\ z^2 - 367 z & = 12 y^2 + 372 y, & (2) \end{align*} as a function of the parameter $y \in \mathbb{R}$. Note that the discriminant of (1) and (2) are strictly positive and we find the following solutions \begin{align*} x_1(y) &= \cfrac{1126 + \sqrt{1126^2 + 4(32 y^2 + 992 y)}}{2},\\ x_2(y) &= \cfrac{1126 - \sqrt{1126^2 + 4(32 y^2 + 992 y)}}{2}, \\ z_1(y) &= \cfrac{367+ \sqrt{367^2 + 4(12 y^2 + 372 y)}}{2},\\ z_2(y) &= \cfrac{367 - \sqrt{367^2 + 4(12 y^2 + 372 y)}}{2}. \end{align*} Thus, we need to study the intersections of $x_1$ with $z_1$ and $z_2$ and the ones of $x_2$ with $z_1$ and $z_2$. We note that $x_1$ does not intersect $z_2$ since the minimum of $x_1$ is higher than the maximum of $z_2$. Moreover $x_1$ and $z_1$ do not intersect as well since it means that $$ \underbrace{(1126 - 367)}_{(1)} + \underbrace{\big(\sqrt{1126^2 + 4(32 y^2 + 992 y)} - \sqrt{367^2 + 4(12 y^2 + 372 y)}\big)}_{(2)} = 0. $$ This is not possible because (1) and (2) are always positive. Additionally, $x_2$ and $z_1$ do not intersect since the minimum of $z_1$ is again higher than the maximum of $x_2$. It remains to study the intersections of $x_2$ and $z_2$. It consists in solving a fourth order equation and I found the follwing values for $y$ $$ y = 0,\; y = -31,\; y = -31/2 - \sqrt{359503}/10,\; y = -31/2 + \sqrt{359503}/10. $$

Answer 2

These equations describe the intersection of two quadric surfaces (hyperbolic cylindres), which is a complicated curve. You can parameterize it by writing

$$3x^2-3(1126)x=96y^2+24(124)y=8z^2-4(734)z=t$$

and solving $x,y,z$ in terms of $t$. Note that there will be eight arcs. $t$ must be larger than the highest minimum, which is that of $96y^2+24(124)y$.

There are eight easy points with $t=0$.