6
$\begingroup$

If $x,y,z\in {\mathbb R}$, Solve this system equation:

$$ \left\lbrace\begin{array}{ccccccl} x^4 & + & y^2 & + & 4 & = & 5yz \\[1mm] y^{4} & + & z^{2} & + & 4 & = &5zx \\[1mm] z^{4} & + & x^{2} & + & 4 & = & 5xy \end{array}\right. $$

This is an olympiad question in Turkey (not international), which that, I could not solve it.

My idea:

$xy=a \\ yz=b \\ xz=c$

$$ \left\lbrace\begin{array}{ccccccl} a^2c^3 & + & b^3a & + & 4b^2c & = & 5b^3c \\[1mm] a^{3}b^2 & + & bc^3 & + & 4c^2a & = &5c^3a \\[1mm] b^{3}c^2 & + & a^3c & + & 4a^2b & = & 5a^3b \end{array}\right. $$

Yes, I know, this is a stupid idea, because it did not work at all ( last system equation is more difficult).

$\endgroup$
  • 1
    $\begingroup$ Symbolic computation reveals that the only solution is $x=y=z=\sqrt2$. $\endgroup$ – Parcly Taxel Dec 8 '17 at 12:00
  • $\begingroup$ Is this really a difficult equation? $\endgroup$ – MathLover Dec 8 '17 at 12:31
  • $\begingroup$ @MathLover If you are ok, you can set as solved. Thanks! $\endgroup$ – user Dec 9 '17 at 9:41
3
$\begingroup$

I checked the Possible Solutions thoroughly. The following solution is possible:

$$x^4+y^2+4+y^4+z^2+4+z^4+x^2+4-5yz-5xz-5xy=0 \Rightarrow (x^4-4x^2+4)+(y^4-4y^2+4)+(z^4-4z^2+4)+\left(\frac {5x^2}{2}-5xy+\frac {5y^2}{2} \right)+\left(\frac {5x^2}{2}-5xz+\frac {5z^2}{2} \right)+\left(\frac {5y^2}{2}-5yz+\frac {5z^2}{2} \right)=0\Rightarrow (x^2-2)^2+(y^2-2)^2+(z^2-2)^2+\left(x \sqrt{\frac 52}-y \sqrt{\frac 52}\right)^2+ \left(x \sqrt{\frac 52}-z \sqrt{\frac 52}\right)^2+ \left(y \sqrt{\frac 52}-z \sqrt{\frac 52}\right)^2=0$$

You can continue from here.

Only solutions are $x=y=z=±\sqrt2$

$\endgroup$
  • $\begingroup$ Good. Your last equation shows $x=y=z$. If we could argue that (by symmetry?), then we could just solve $x^4+x^2+4 = 5x^2$. $\endgroup$ – B. Goddard Dec 8 '17 at 12:54
  • $\begingroup$ Are you talking about complex solutions? $\endgroup$ – Zaharyas Dec 8 '17 at 13:03
  • $\begingroup$ No. You get $(x^2-1)^2 = 0$ which has only real solutions. $\endgroup$ – B. Goddard Dec 8 '17 at 13:07
2
$\begingroup$

Summing up all the equtions you find:

$$\sum_{cyc} x_i^4 + \sum_{cyc} x_i^2 - 5\sum_{cyc} x_ix_j +12= 0$$

Now manipulate in such way to have the equality as sum of squares!

EG

$$-5xy =\frac52 \left( x-y\right)^2-\frac52x^2-\frac52y^2$$

That is:

$$x^4+y^4+z^4-4x^2-4y^2-4z^2+\frac52 \left( x-y\right)^2+\frac52 \left(y-z\right)^2+\frac52 \left( z-x\right)^2+12=0$$

$$(x^2-2)^2+(y^2-2)^2+(z^2-2)^2+\frac52 \left( x-y\right)^2+\frac52 \left(y-z\right)^2+\frac52 \left( z-x\right)^2=0$$

The system has 2 different solutions:

$$x=y=z=\sqrt 2$$

and

$$x=y=z=-\sqrt 2$$

$\endgroup$
1
$\begingroup$

Your system can be written as follows

$$ \left\lbrace\begin{array}{clc}\tag{1} \left( {x}^{2}-2 \right) ^{2}+ \left( y-z \right) ^{2}+ \left( x-z \right) \left( x+z \right) +3\,({x}^{2}-\,zy)&=&0 \\ \\[1mm] \left( {y}^{2}-2 \right) ^{2}+ \left( z-x \right) ^{2}+ \left( y-x \right) \left( y+x \right) +3(\,{y}^{2}-\,zx)&=&0 \\ \\[1mm] \left( {z}^{2}-2 \right) ^{2}+ \left( x-y \right) ^{2}+ \left( z-y \right) \left( z+y \right) +3(\,{z}^{2}-\,xy) &=&0 \end{array}\right. $$

Now one solution of the $(1)$ is $x=y=z=\sqrt{2}$ which is mentioned in the first comment.

$\endgroup$
0
$\begingroup$

Another way, adding gives you $$\sum x^4 + \sum x^2 +4 =5\sum xy$$ where $\sum$ represents cyclic sums. However by AM-GM, $\sum (x^4+4) \geqslant 4\sum x^2$ and $\sum x^2 \ge \sum xy$.

So we need equality in both those inequalities, which is possible iff $a=b=c=\pm\sqrt2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.