4. On the coordinate plane, an isosceles triangle $A B C$ was drawn: $A B=2016, B C=$ $A C=1533$, with vertices $A$ and $B$ lying on nodes on the same horizontal line. Determine how many nodes lie within the triangle $A B C$ (including nodes lying on the sides). A node is a point on the coordinate plane where both coordinates are integers.
Solution. Note that $1533^{2}-1008^{2}=(1533-1008)(1533+1008)=525 \cdot 2541=21 \cdot 25 \cdot 7 \cdot 363=$ $7 \cdot 3 \cdot 5^{2} \cdot 7 \cdot 3 \cdot 11^{2}=(7 \cdot 3 \cdot 5 \cdot 11)^{2}=1155^{2}$. Therefore, the height of the triangle is 1155.

We see that the GCD of 1155 and 1008 is 21. This means that there are 22 nodes (including the vertices) on the side, which divide it into 21 equal parts.

The further part of the solution is based on the fact that the triangle gradually transforms into a rectangle $1008 \times 1155$, the number of nodes in which is easy to calculate. This process is schematically shown in the figure. Let's denote the required number of nodes in the triangle by $T$.
![](https://cdn.mathpix.com/cropped/2024_05_06_46c72bd51b2f9098b814g-1.jpg?height=350&width=956&top_left_y=1766&top_left_x=542)

1) Cut the triangle along the axis of symmetry into two halves (two right triangles) and consider each of them as a separate figure. The sum of the number of nodes in these halves is 1156 more than in the original triangle, that is, $T+1156$, since the nodes that were on the axis of symmetry have been duplicated.
2) Combine these two halves into a rectangle $1008 \times 1155$. The inner parts of the halves completely cover the rectangle, while 22 nodes, located in these halves, coincide. Therefore, the number of nodes in the rectangle is 22 less than in the two halves, that is, $T+1156-22$. At the same time, this number is obviously equal to $1009 \cdot 1156$. Thus, $T+1156-22=1009 \cdot 1156$; $T=1008 \cdot 1156+22=1156000+9248+22=1165270$.

Answer: 1165270 nodes.