Difficult system of 3 equations

[Math Tutor]

[tex](x^{2}-6x+13)y = 20\\ (y^{2}-6y+13)z = 20\\ (z^{2}-6z+13)x = 20[/tex]

Find all real solutions of the system.

[Guest]

Since [tex]t^2-6t+13>0[/tex] for any t, [tex]x, y, z[/tex] must be positive.
[tex]\frac{x^2-6x+13}{x}.\frac{y}{z^2-6z+13}=1[/tex]
[tex](x+\frac{13}{x}-6).\frac{y}{z^2-6z+13}=1[/tex], division by the second equation leads to
[tex](x+\frac{13}{x}-6).\frac{y}{y^2-6y+13}.\frac{z}{z^2-6z+13}=\frac{y^2}{20}[/tex]
[tex](x+\frac{13}{x}-6).\frac{1}{y+\frac{13}{y}-6}.\frac{1}{z+\frac{13}{z}}=\frac{y^2}{20}[/tex].
Let [tex]x+\frac{13}{x}-6=a[/tex], [tex]y+\frac{13}{y}-6=b[/tex], [tex]z+\frac{13}{z}-6=c[/tex]. All of them are positive.
[tex]\frac{bc}{a}=\frac{20}{y^2}[/tex]. Analogically, we have
[tex]\begin{tabular}{|l}y^2bc=20a\\z^2ac=20b \\x^2ab=20c \end{tabular}[/tex]
Therefore
[tex]\begin{tabular}{|l}y^2b^2=a^2z^2 \\ z^2c^2=b^2x^2 \\ x^2a^2=c^2y^2\end{tabular}[/tex]
We square root:
[tex]\begin{tabular}{|l}by=az \\ cz=bx \\ ax=cy \end{tabular}[/tex]
Take the last one and multiply by y:
[tex]axy=cy^2[/tex], but [tex]axy=(x^2-6+13)y=20[/tex], therefore [tex]cy^2=20[/tex] Substituting into the first equation of the first system, namely [tex]cy^2b=20a[/tex] we get [tex]a=b[/tex]. Symetrically, [tex]a=b=c[/tex]. This implies that at least two of [tex]x,y,z[/tex] are equal. WLOG [tex]x=y[/tex], the first equation becomes
[tex]x^3-6x^2+13x-20=0[/tex] with a solution [tex]x=y=4[/tex]. Substituting into the next equation, we get z=4.
Answer: [tex](4;4;4)[/tex].

[Guest]

I like the solution but wonder if there is something more simple.
Seems that the solution is quite complicated for a system with 3 equations.
I must confess that I does not find something easier but it should exist.

[Math Tutor]

The system is solved.
Here is the next problem of the month.
viewtopic.php?f=42&t=635
Again system of 2 equations

[Guest]

The proof given above is nonsense [tex]\frac{y^2}{20}[/tex] should be [tex]\frac{z^2}{20}[/tex] in the fourth line which causes the whole thing to break down.

Below is a proper proof:

Spoiler: show

Rearranging the first equation gives [tex]y = 20/((x-3)^2+4)[/tex]. Observe that [tex]x\in (-\infty,+\infty)[/tex] implies [tex]y\in(0,5][/tex] from considering the image of [tex]y[/tex] in [tex]y = 20/(x^2-6x+13)[/tex]. By symmetry this means [tex]x\in (0,5][/tex]. Applying the same trick we see that [tex]x\in (0,5][/tex] implies [tex]y\in(\frac{20}{13},5][/tex], and by symmetry [tex]x\in (\frac{20}{13},5][/tex]. Repeating one last time tells us that [tex]x,y,z\in [\frac{5}{2},5][/tex].

Now notice that if [tex]x\in [\frac{5}{2},4][/tex] then [tex]y\in[4,5][/tex] and if [tex]x\in [4,5][/tex] then [tex]y\in[\frac{5}{2},4][/tex]. We get similar results for [tex]y[/tex] and [tex]z[/tex].

So either
[tex]x\in [\frac{5}{2},4]\Rightarrow y\in[4,5]\Rightarrow z\in [\frac{5}{2},4]\Rightarrow x\in[4,5][/tex]
or
[tex]x\in[4,5]\Rightarrow y\in [\frac{5}{2},4]\Rightarrow z\in[4,5]\Rightarrow x\in [\frac{5}{2},4].[/tex]
In either case [tex]x[/tex] is in both [tex][\frac{5}{2},4][/tex] and [tex][4,5][/tex], so must be [tex]4[/tex]. This implies the only solution is [tex]x=y=z=4[/tex].

Hope this helped,

R. Baber.

[leesajohnson]

This one is difficult to solve and it's solution is so lengthy.