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

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).

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

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 $x=y=z=±\sqrt2$

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

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

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

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.

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 $a=b=c=\pm\sqrt2$.