3

This was a high school entrance exam problem.

Find all real solutions for $x,y,z$: $$\begin{cases}x+y^2+z^3=3\\y+z^2+x^3=3\\z+x^2+y^3=3\end{cases}$$

I believe the intended solution uses basic algebra only, but it's allowed to use higher level maths such as calculus, elliptic curves or 3d geometry if we can.

My attempt: I guess $(1,1,1)$ is the only solution, so I tried substituting $x=x_1+1$ and similar for $y$ and $z$, but to no avail.

The usual way of solving a system of equations yields

$$x+y^2=3+(x^2+y^3-3)^3$$ $$x^3+y=3-(x^2+y^3-3)^2$$

and I can't proceed further.

Saturday
  • 1,358
  • You can graph $f(x,y)=0$ on some online graphing calculators – Chris Sanders Mar 13 '22 at 06:29
  • 1
    Btw which entrace exam is this?,..just curious.. –  Mar 13 '22 at 06:29
  • The resultant of your two equations is a single polynomial in y, of degree 27. Its only real root seems to be $y=1$ according to Wolfram Alpha – Empy2 Mar 13 '22 at 07:23
  • The vector $(x-y,y-z,z-x)$ is orthogonal to $1,y+z,z^2+zx+x^2)$ and two similar vectors. Perhaps you can show they are linearly independent, with a nonzero triple product. Then the first vector must be zero? – Empy2 Mar 13 '22 at 07:30
  • This question has another solution but only for positive reals. – Toby Mak Mar 13 '22 at 08:54

0 Answers0