If we consider two parabolas $y^2=4ax$ and other one $x^2=4ay$. Both parabolas meet at origin and some other point for reference consider $(x_1,y_1)$ .Now we have to find number of integral points between the enclosed part which is between $(0,0)$ and $(x_1,y_1)$. How to pursue finding it? From my point of view I would draw constant lines in the region such as $y=1$, $y=2$, $y=3$ or $x=1$, $x=2$, $x=3$ and try to find symmetry between the points But this won't help instead of making the question lengthy. So what can be the other useful approach to this?
Thanks
As shown in $\mathrm{Fig.\space 1}$, two parabolae $x^2=4ay$ and $y^2=4ax$ intersect at $O_1\space (0,0)$ and $O_2\space (4a,4a)$. The region bounded by this pair of parabolae is symmetrical about the line $O_1O_2$, the equation of which is $y=x$. Therefore, in order to find a formula that gives the number of lattice points inside the mentioned region, we need to find the number of points that lie within only one half of it. Please note that all lattice points under consideration lie on lines, which are parallel to $O_1O_2$ and pass through the integral points on the $y-$axis
It can be also shown that the line $y=x+a$, which is parallel to $O_1O_2$, is tangent to the parabola $y^2=4ax$ at $D_1\space (a,2a)$. Therefore, the line parallel to $O_1O_2$, which passes thorough $C\space (0,\lfloor a\rfloor)$ is the furthest line from $O_1O_2$, which also goes through a set of lattice points. Our aim is to find the lattice points on all lines such as $y=x+m$, where $m$ is an integer and $0\le m\le\lfloor a\rfloor$.
The line $y=x+m$ intersects the parabola $y^2=4ax$ at $B_1$ and $B_2$. Their coordinates are as follows; $$B_1 = (2a-m-2\sqrt{a}\sqrt{a-m},\space\space 2a-2\sqrt{a}\sqrt{a-m}), \quad\text{where}\quad a\ge m $$ $$B_2 = (2a-m+2\sqrt{a}\sqrt{a-m},\space\space 2a+2\sqrt{a}\sqrt{a-m}), \quad\text{where}\quad a\ge m.$$
If $\bar{B}_1$ and $\bar{B}_2$ are the two nearest lattice points to $B_1$ and $B_2$ respectively, we can express their coordinates as shown below. As the previous pair of expressions, they are also valid if and only if $a\ge m$. $$\bar{B}_1 = (\lceil 2a-m-2\sqrt{a}\sqrt{a-m}\rceil,\space\space \lceil 2a-2\sqrt{a}\sqrt{a-m}\rceil)$$ $$\bar{B}_2 = (\lfloor 2a-m+2\sqrt{a}\sqrt{a-m}\rfloor,\space\space \lfloor 2a+2\sqrt{a}\sqrt{a-m}\rfloor)$$
Number of lattice points $n_m$ lying on $B_1B_2$ is one more than the difference between either $x-$ or $y-$coordinates of the lattice points $\bar{B}_1$ and $\bar{B}_2$. Therefore, we shall write, $$n_m = \lfloor 2a+2\sqrt{a}\sqrt{a-m}\rfloor - \lceil 2a-2\sqrt{a}\sqrt{a-m}\rceil +1.\tag{1}$$
In a similar vein, we can express the number of lattice points $n_0$ located on the line of symmetry $O_1O_2$ as, $$n_0 = \lfloor 4a\rfloor - \lceil 0\rceil +1 = \lfloor 4a\rfloor +1.\tag{2}$$
Now, we are in a position to write down the formula for the total number of lattice point $N$ inside and on the border of the region in question. $$N=n_0+2\sum_{m=1}^{\lfloor a\rfloor} n_m \tag{3}$$
When we substitute (1) and (2) in (3), we get, $${\small{\text{For}\space a\ge 1}}\quad N=\lfloor 4a\rfloor +1+2\left( \lfloor a\rfloor +\sum_{m=1}^{\lfloor a\rfloor} \left(\lfloor 2a+2\sqrt{a}\sqrt{a-m}\rfloor-\lceil 2a-2\sqrt{a}\sqrt{a-m}\rceil \right)\right),$$ $\space{\small{\text{For}\space a\le 1}}\quad N=\lfloor 4a\rfloor +1.$
As shown in $\mathrm{Fig.\space 2}$, the graph of Number-of-Lattice-Point vs $a$ and its second order polynomial trend curve are hugging each other so closely, we see only one of them. This is no wonder, because the measure of goodness-of-fit $R^2$ is equal to 1. It also improves the estimate given by @JeanMarie in one of his comments from $N\approx 5.33a^2$ to $N\approx 5.3447a^2$.
Why is the number of lattice points one more than the difference between the $x−$ or $y−$coordintes of the last and the first lattice ponts?
Let us imagine a straight fence erected to divide two plots of land. This fence has $\space\large{n}\space$ number of $\large{\text{1-meter}}$-spaced upright poles and the poles are numbered from $\large{1}$ to $\large{n}$. If we assume that the number given to each pole is either its $x-$ or $y-$coordinate, then each number represents a grid point along the chosen axis. Since there are $\large{n}$ number of poles, there are $\large{n}$ number of grid points. Now, if you subtract the number or the coordinate given to the last pole (i.e., $\large{n}$ ) from the number or the coordinate given to the first pole (i.e., $\large{1}$ ), you get $\large{n-1}$ as their difference. But, there are $\large{n}$ number of poles ( or grids points), which is one more than the difference.