Thee number of points with integral coordinates that lie in the interior of the region common to the circle $x^2+y^2=16$ and the parabola $y^2=4x$ is

Question

the number of points with integral coordinates that lie in the interior of the region common to the circle $x^2+y^2=16$ and the parabola $y^2=4x$ is

Answer 1

$$x^2+y^2\leq16$$ and $$x\geq\frac{y^2}{4}$$ give $$\frac{y^4}{16}+y^2\leq16$$ or $$-\sqrt{\sqrt{320}-8}\leq y\leq \sqrt{\sqrt{320}-8},$$ which gives $$-3\leq y\leq 3$$ and a smooth checking.

I got $19$ points.

Answer 2

The admssible $x$-values of such points are $1$, $2$, $3$. The circle and the parabola intersect at the two points $(2.47,\pm 3.14)$. When $x<2.47$ the admissible $y$-values have to satisfy $y^2<4x$, and when $x>2.47$ they have to satisfy $y^2<16-x^2$. Therefore we obtain the points $$(1,-1), (1,0),(1,1)\quad /\quad (2,-2),\ldots, (2,2)\quad /\quad (3,-2),\ldots, (3,2)\ ,$$ which are $13$ points in all.

Answer 3

We must have $0<x<4$ (assuming we don't count points on either curve) so $x=1,2,3$.

For each $x$, take all $y$ such that $y^2<\min\{4x,16-x^2\}$:

$x=1$ $\implies$ $y^2<\min\{4,15\}=4$ $\implies$ $y=0,\pm1$.
$x=2$ $\implies$ $y^2<\min\{8,12\}=8$ $\implies$ $y=0,\pm1,\pm2$.
$x=3$ $\implies$ $y^2<\min\{12,7\}=7$ $\implies$ $y=0,\pm1,\pm2$.

So there are 13 points altogether. (Or 8 if you consider just $y\ge0$.)