What is the smallest sphere whose surface includes 100 integer points? Let $S(r)$ be the surface of the origin-centered sphere in $\mathbb{R}^3$.
A point is an integer point if all its coordinates are integers.

What is the smallest radius $r_n$ such that $S(r_n)$ includes $\ge n$ integer points?

What is the growth rate of $r_n$ with respect to $n$?
Is there an algorithm that could compute $r_n$ for a specific given $n$?
It is known that rational points (all coordinates rational) are dense on $S(1)$:
see, e.g., the MO question
Rational points on a sphere in $\mathbb{R}^d$.
One possible approach is via rational points of bounded height.
The height of a rational $a/b$ in lowest terms is max$(|a|,|b|)$, and
the height of a rational point is the max of the heights of its coordinates.

          


          

Rationals of height $\le 2048$ on sphere.
Image due to Stefan Kohl in this answer.


Choose an appropriate $h_\max$, perhaps using an
estimate
of the number of rational points of height at most $h_\max$ on $S(1)$.
Then scale all coordinates by the LCM of the points' denominators.
For example, for $h_\max=10$, scaling by $2^3 \cdot 3^2 \cdot 5 \cdot 7 = 2520$
would suffice to clear all denominators,
so that $S(2520)$ includes all those points at integer coordinates.

But this would not necessarily result in the minimum $r_n$ for a given $n$.
It would likely be better to use rational points that result in a small LCM.
Exact calculations on a sphere 
(for example, computing Voronoi diagrams on a sphere) often need integer points
of bounded size.

Update. The exact answer to the question in the title,
due to Dap and Gerhard Paseman, is
that a sphere of radius $\sqrt{74} \approx 8.6$ includes $120$ integer points
on its surface,
and smaller spheres include fewer than $100$ points.
 A: Indeed I was thinking too linearly.  If we change standard notation slightly, and call $r_3(n)$ the number of (unordered) representations of $n$ as a sum of three increasing and distinct squares, then a sphere with squared radius of $n$ and centered at the origin has at least $48r_3(n)$ integer lattice points on it.  One also has contributions when two of the three squares are equal, or one or more of the squares is zero. We have $r_3(74)\geq 2$, as well as $74=49+25$, so this gives at least (and exactly) $120$ points on this sphere of radius less than 9.  Thanks to Dap for finding a relevant OEIS sequence.  I leave the growth rate of $r_3$ buried in the literature, but similar quantities such as (ordered) tuples counted by $r_4$ are well documented under sums of squares.
Gerhard "Then There Are Off-Origin Spheres" Paseman, 2018.01.30.
A: I will just reference this link Many representations as a sum of three squares 
that shows that this is a really hard questions and seems open at the moment to estimate the growth of the sequence.
A: Here are a few facts about this problem, quoting mostly from Local statistics of lattice points on the sphere by Jean Bourgain, Peter Sarnak, Zeév Rudnick:
''A celebrated result of Legendre/Gauss asserts that $n$ is a sum of three squares if and only if $n\ne 4^a(8k+7)$. Let 
$$ N(n) = \bigl\{ (x,y,z)\in\mathbb Z^3 : x^2+y^2+z^2=n \bigr\}. $$
The  behaviour  of $N(n)$ is  very  subtle,  and  it  was  a  fine  achievement  in  the 1930’s  when  it  was  shown  that $N(n)\to\infty$ as $n\to\infty$ through square-free values of $n$. It is known that $N(n)\ll n^{1/2+o(1)}$, and if there are primitive solutions, i.e., with $\gcd(x,y,z)=1$, which happens if and only if $n\not\equiv 0,4,7\pmod8$, then there is a lower bound $N(n)\gg n^{1/2-o(1)}$. This lower bound is ineffective and indicates that the behaviour of $N(n)$ is still far from being understood.''
