You don't have to loop through all coordinates to check if two squares intersect.
Here's a simple solution which will work as long as the squares are not rotated.
Say that you represent a square by its top left corner coordinate and its side length. Let aX and aY represent the coordinates and aLen the side length of square A, and vice versa for square B.
Then to check if square B intersects with square A evaluate this:
(aX < (bX + bLen) && (aX + aLen) > bX)
&& (aY < (bY - bLen) && (aY - aLen) > bY)
In other words, there are four possible scenarios and you check if either of square B's corners are within the X-range of square A and that either of square B's corners are within the Y-range of square A.
(Y)
 ^
 |             1:                        2:
 |       +--------+                   +--------+
 |       |        |                   |        |
 |       |   A +--|-----+       +-----+--+ A   |
 |       |     |  |     |       |     |  |     |
 |       +-----+--+ B   |       |   B +--+-----+
 |             |        |       |        |
 |             +--------+       +--------+
 |
 |             3:                        4:
 |       +--------+                   +--------+
 |       |        |                   |        |
 |       |   B +--|-----+       +-----+--+ B   |
 |       |     |  |     |       |     |  |     |
 |       +-----+--+ A   |       |   A +--+-----+
 |             |        |       |        |
 |             +--------+       +--------+
 |
 +-------------------------------------------------> (X)
For more info see this answer of a similar question: Determine if two rectangles overlap each other?