<s>[INST] 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.
 [/INST] It seems that solving it for $y$ is a better idea, as we are going to have $y=5\pm\sqrt{something}$.
Let us solve it for a number with any number of digits: $\overline{...vwxyz}$.
$y^2-10y+(z^2-z)+(x^2-100x)+(w^2-1000w)+...=0$
$y=5\pm\sqrt{25-(z^2-z)-(x^2-100x)-(w^2-1000w)-...}$.
Note, that $0\le y\le9$, therefore, $0\le(z^2-z)+(x^2-100x)+(w^2-1000w)+...\le25$.
But $z^2-z\le9^2-9=72$, while $x^2-100x\le1^2-100=-99$ for $1\le x\le9$, $w^2-1000w\le-999$ for $1\le w\le9$ etc.
Hence, $x=w=...=0$ and $25-z(z-1)$ is a square. For $z=0,1,2,3,4,5$ we get $25,25,23,19,13,5$ and only the first two are squares.
The only two numbers are $0$ and $1$.
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