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

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]

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Why not title this something like "Are there always nontrivial real solutions to $A_{1}x^{5}+B_{1}y^{5}+C_{1}z^{5}=0$ and $A_{2}x+B_{2}y+C_{2}z=0$ ?" – j.c. Jul 8 '10 at 2:09
jc, yeah, your title would be better. – user4606 Jul 8 '10 at 2:13
You can edit the title of your question. – j.c. Jul 8 '10 at 2:14
Joel, your are right. It would be simpler to state the question in your way. – user4606 Jul 8 '10 at 2:25
I changed the title per jc's suggestion. – Theo Johnson-Freyd Jul 14 '10 at 6:01

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 $y$):

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jc, thank you for the answer. It looks very much like an AG question, but I think there should be an explanation without using AG results or AG language. – user4606 Jul 8 '10 at 2:25
Ok, let's make it simple. As said before, plug in $x$, get a degree 5 homogenous expression $f$ in two variables $y$ and $z$. Expand this expression. Plug in $y = 1$, if the degree of the equation drops below 5, then every term of $f$ contained a $y$, so $y = 0, x =$ anything'' is a solution. Otherwise, you have a degree 5 equation in $z$ so it has at least one real solution. – Karl Schwede Jul 8 '10 at 2:38
Thanks Karl, you just beat me to it. I'll just add that it follows for suitably generic choices of your coefficients that the solution sets (x,y,z) will be 1, 3 or 5 straight lines in $R^3$ passing through the origin. – j.c. Jul 8 '10 at 2:42
Karl, thanks for your answer. It would be highly appreciated if you could help on the second question just added. – user4606 Jul 8 '10 at 2:50
Hi, unknown, I noticed that you haven't been voting up answers to any of your questions. You should consider doing so if you find them helpful. – j.c. Jul 8 '10 at 3:04

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).

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Qiaochu, thanks for the answer. It would be highly appreciated if you could also help on the second question just added. – user4606 Jul 8 '10 at 2:55