Number of points $C=(x,y)$ on the circle $x^2+y^2=16$ such that the area of the triangle whose vertices are $A,B$ and $C$ is a positive integer Let $A(-4,0)$ and $B(4,0)$.Number of points $C=(x,y)$ on the circle $x^2+y^2=16$ such that the area of the triangle whose vertices are $A,B$ and $C$ is a positive integer,is...

I found the area of triangle $ABC=4|y|$ but i cannot count in how many cases the area will be a positive integer.The answer given is $62$.Is there a method by which i can solve this.Please help me.Thanks.

Hint: $-4 \leq y \leq 4$, so the possible $y$ values are $$\frac{1}{4}, -\frac{1}{4}, \frac{2}{4}, -\frac{2}{4}, \frac{3}{4}, - \frac{3}{4}, \dots, \frac{16}{4}, -\frac{16}{4}$$
Then, how many times is each $y$ value realized on the circle?
