Integers on a sphere

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Consider a point $(a,b,c)$ on the sphere, as shown in the diagram. 

The yellow and green triangles are both right-angled, and the hypotenuse of the yellow triangle is one of the sides of the green triangle. Let this side be $k$. All of the other sides of the triangles are known in terms of $a$, $b$ and $c$, so we can make an expression linking $a$, $b$ and $c$ by applying Pythagoras' Theorem to the two triangles.

Yellow triangle: $a^2+b^2=k^2$

Green triangle: $k^2+c^2=3^2\Rightarrow a^2+b^2+c^2=9$

Since $a$, $b$ and $c$ are integers, $a^2$, $b^2$ and $c^2$ must be $0$, $1$, $4$ or $9$. Choosing $a^2$ first, and then $b^2$ so that $a^2+b^2\le9$, and then $c^2$ so that $a^2+b^2+c^2=9$, the possible options are shown in this table.

$a^2$	$b^2$	$c^2$
$0$	$0$ $1$ $4$ $9$	$9$  $8$ not possible $5$ not possible $0$
$1$	$0$ $1$ $4$	$8$ not possible $7$ not possible $4$
$4$	$0$ $1$ $4$	$5$ not possible $4$ $1$
$9$	$0$	$0$

  

That gives $6$ possible options for $a^2$, $b^2$ and $c^2$.

For $0,0,9$ and so on, $a$, $b$ and $c$ are $0$, $0$ and either $+3$ or $-3$ in some order. So for the $3$ options that are made up of two $0$s and a $9$, there are $3\times2=6$ integer points on the sphere.

For an option such as $4$, $4$, $1$, we could have $(2,2,1)$, $(2,2,-1)$, $(2,-2,1),$ $(2,-2,-1)$, $(-2,2,1)$, $(-2,2,-1)$, $(-2,-2,1),$ $(-2,-2,-1)$ - which gives $8$ integer points. So for the $3$ options that are made up of two $4$s and a $1$, there are $3\times8=24$ integer points on the sphere.

That gives a total of $6+24=30$ integer points on the sphere.