# Integer ordered pairs in logarithmic equation

#### jacks

##### Well-known member
no. of integer ordered pairs of $(x,y,z)$ in

$\sqrt{x^2-2x+6}\cdot\log_{3}(6-y) = x$

$\sqrt{y^2-2y+6}\cdot\log_{3}(6-z) = y$

$\sqrt{z^2-2z+6}\cdot\log_{3}(6-x) = z$

My approach :: Here $6-x,6-y,6-z>0$. So $x,y,z<6$

Now $\displaystyle \log_{3}(6-y) = \frac{x}{\sqrt{x^2-2x+6}}=\frac{x}{\sqrt{(x-1)^2+5}}$

and $\displaystyle \log_{3}(6-z) = \frac{y}{\sqrt{y^2-2y+6}}=\frac{y}{\sqrt{(y-1)^2+5}}$

and $\displaystyle \log_{3}(6-x) = \frac{z}{\sqrt{z^2-2z+6}}=\frac{z}{\sqrt{(z-1)^2+5}}$

How can I calculate after that.

I don't have a reference for this, but I guess that if $n$ is an integer then $\log_3n$ is transcendental unless $n$ is a power of $3$. If so, then the only solution to those equations must be the obvious one $x=y=z=3$.