Are there always nontrivial real solutions to $A_1 x^5 + B_1 y^5 + C_1 z^5 = 0$ and $A_2 x + B_2 y + C_2 z = 0$? - MathOverflow

Are there always nontrivial real solutions to $A_1 x^5 + B_1 y^5 + C_1 z^5 = 0$ and $A_2 x + B_2 y + C_2 z = 0$?

2010-07-08T01:52:16Z

Firstly, are there always nontrivial real solutions to the sysytem of equations, $A_{1}x^{5}+B_{1}y^{5}+C_{1}z^{5}=0$ and $A_{2}x+B_{2}y+C_{2}z=0$, for real numbers $A_{1}$, $B_{1}$, $C_{1}$, $A_{2}$, $B_{2}$, and $C_{2}$? [Answered]

Secondly, are there always nontrivial real solutions to the sysytem of equations, $A_{1}\frac{x^{5}}{\left|x\right|}+B_{1}\frac{y^{5}}{\left|y\right|}+C_{1}\frac{z^{5}}{\left|z\right|}=0$ and $A_{2}x+B_{2}y+C_{2}z=0$, for real numbers $A_{1}$, $B_{1}$, $C_{1}$, $A_{2}$, $B_{2}$, and $C_{2}$? [Answered]

Answer by jc for Are there always nontrivial real solutions to $A_1 x^5 + B_1 y^5 + C_1 z^5 = 0$ and $A_2 x + B_2 y + C_2 z = 0$?

2010-07-08T02:13:34Z

Without loss of generality, let $A_2\neq0$. Then we have $x=-\frac{B_2}{A_2}y-\frac{C_2}{A_2}z$. Thus we can eliminate $x$ from the first equation to get a degree 5 equation describing a plane curve in $y$ and $z$.

~~By Harnack's curve theorem, the number of components of this curve in the real projective plane is between 1 and 7. Thus you should expect at least one family of solutions to your equations.~~ This paragraph was nonsense because what one really has at this point is an equation relating points on the projective line.

See Karl Schwede's comment or Qiaochu Yuan's answer for a correct characterization.

Note that their arguments extend to the second question as well. Again, we can look at the $y=1$ slice of the equation we get on eliminating $x$, something like:

$1+C_1\frac{z^5}{|z|}+\frac{(-B_2-C_2z)^5}{|-B_2-C_2z|}=0$

where I've scaled out $A_1,A_2,B_1$. For large positive $z$ and large negative $z$ the function on the left will take opposite signs (which sign is taken will depend on the signs and relative magnitudes of $C_1$ and $C_2$), so you must have at least one root in between.

In the first question, your solutions typically end up as lines through the origin because the homogeneous equation in $y$ and $z$ can be rewritten as one for $y/z$. This doesn't work in your second question and you get much more interesting looking curves (the ordinate is $z$ and the abscissa is $y$):

Answer by Qiaochu Yuan for Are there always nontrivial real solutions to $A_1 x^5 + B_1 y^5 + C_1 z^5 = 0$ and $A_2 x + B_2 y + C_2 z = 0$?

2010-07-08T02:36:54Z

Yes. Suppose otherwise and let $\mathbf{u} = (u_1, u_2, u_3)$ and $\mathbf{v} = (v_1, v_2, v_3)$ be two linearly independent points on the hyperplane which do not intersect the quintic. Then some point of the form $\mathbf{u} + \mathbf{v} t$ must intersect the quintic because the corresponding polynomial in $t$ is of degree $5$ (in particular, its leading and constant coefficients are nonzero).