Find a three digit number ($\overline{xyz}$)? (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.

Since $x^2+y^2+z^2\leq 3\times 9^2 = 243$, we know that $x=0$, $x=1$, or $x=2$. But if $x=2$, then $x^2+y^2+z^2 \leq 2\times 9^2 + 4 = 166$, so we must definitely have $x=0$ or $x=1$.
But $x=1$ doesn't work: that would give us a maximum possible total of $1+9^2+9^2 = 163$; that would meant that $y\leq 6$; but then the best we can do is $1+6^2+9^2 = 118$, which means $y\leq 1$, but then we can't even get to $100$, since the maximum possible total would be $1+1+81$.
So we must have $x=0$.
Now we are down to $y^2+z^2 = 10y+z$, or $y^2 - 10y + (z^2-z) = 0$. 
This means $y = 5\pm \frac{\sqrt{100 - 4(z^2-z)}}{2}$. So $100-4(z^2-z)$ must be an even square. So $z^2-z$ must be $0$, $9$, $16$, $21$, $24$, or $25$. 
Since $z^2-z = z(z-1)$, with $0\leq z\leq 9$ integral, this means the total cannot be $9$, $21$, or $25$. If $z\gt 1$, then we cannot have a perfect square, which excludes $16$; and $24$ cannot be written as a product of consecutive integers, which only leaves $z^2-z=0$. Thus, $z=0$ or $z=1$. Both lead to $y=0$ or $y=10$, but the latter is impossible.
So the only possibilities are $000$ and $001$. 
