(Excuse me for my english: I'm spanish speaker)
"Find a three digit number $\overline{xyz}$ such that $x^2 +y^2 + z^2$ is equal to the number (xyz). "
I have this equation: $$x^2 -100x +(y^2-10y+z^2-z)=0$$
And the discriminant (in $x$):
$$\Delta_x = -4y^2 +40y-4z^2+4z+10000$$
Solutions for $x_i$ are: $$x_i = \frac{100 \pm \sqrt{\Delta_x}}{2}$$
in which ${\Delta_x} \geq 0$. The obvious conditions are:
i) $x \neq 0$ and $x=1,..., 9$;
ii) $y, z = 0, ..., 9$.
I need an orientation because pick $(y,z) \in \{0,...9\} \times \{0,...9\}$ I believe it's arduous task.
Another "attemp" from myself is to find $(x, y, z) \in \{1,...9\} \times \{0,...9\} \times \{0,...9\} $ such that
$$(x-50)^2 + (y-5)^2 +\left(z- \frac{1}{2}\right)^2 = \frac{10101}{4}$$ but I'm not skilfull in multivariable calculus :(
Any advice is welcome.