No, It could be copy of Official GMAT question but it belongs to Manhattan GMAT. Anything special with it?

Please don't give tough questions Khs14 is scared

Solution please.

Carcass

Consider the diagram.
The arrows could be vertical, horizontal or diagonal.




The vertical arrows are shown by the blue arrows. 5 of them will start from x = 0, 5 from x = 1 and so on till x = 9. So you have 50 of these blue arrows. You have another 50 vertical arrows which are the same arrows but with the arrow head on the opposite end (shown by the red arrow). So you have a total of 100 vertical arrows.
Similarly, you have 100 horizontal arrows.
Now check out the diagonal arrows. One co-ordinate should be of length 3 and another of 4 (so that the arrow length is 5 and all points are integers). Look at the purple arrows. The x co-ordinate is 3 and the y co-ordinate is 4. You can make 7*6 = 42 such arrows. Similarly, you can make 42 arrows with x cor-ordinate as 4 and y co-ordinate as 3. So you have 84 arrows. But you get another set of 84 arrows by keeping the arrows the same but putting the arrow head on the opposite end so you get a total of 2*84 = 168 arrows.
Similarly, you can make arrows in the opposite direction shown by the green arrows. So you have another 168 arrows.
Total = 100 + 100 + 168 + 168 = 536

Given two points with their co-ordinates: A(x1, y1), B(x2, y2). Then the length of AB should be:

AB^2 = 5^2 =25 = (x1-x2)^2 + (y1-y2)^2 = X^2 + Y^2 in which I denote X = x1-x2 and Y = y1-y2

Our goal is to determine X and Y such that X^2 + Y^2 = 25. Let's find out all the pairs (X^2, Y^2) that meet such a requirement, and also keep in mind that the square root of Y^2 and X^2 has to be integers. One example below does not meet this requirement such as the pair (1,24) as square root of 24 is not an integer:

(X^2,Y^2) = { (0,25), (1,24), ..., (9,16),...} Feel Free to exploit the whole set but we have found only one candidate in this list that meets the requirements above: (9,16) = (3^2, 4^2)

Next, we need to find out all the pairs (x1, x2) that has a difference of 3 and all the pairs (y1,y2) that has a difference of 4.

1. All the pairs (x1, x2) that has a difference of 3: (0,3), (1,4), ..., (6,9) and you flip the coordinates = 14 pairs
AND all the pairs (y1,y2) that has a difference of 4: (0,4), (1,5),...,(5,9) and you flip the coordinates = 12 pairs


2. Question: How many ways for you to choose a pair from 7 pairs AND a pair from 6 pairs? 14 x 12 = 168
Next: Don't forget, there are 2 ways to arrange 2 chose pairs once we choose one from (y1,y2) and one from (x1,x2) so we have a total of 168 x 2 = 336

3. Do the same thing for special case: AB^2 = 5^2 =25 = (x1-x2)^2 + (y1-y2)^2 = X^2 + Y^2 = 0 + 25
Find All the pairs (x1, x2) that has a difference of 0: (0,0), (1,1), ..., (9,9) and you do not flip the coordinates in this case as you will get identical pairs = 10 pairs
AND all the pairs (y1,y2) that has a difference of 5: (0,5), (1,6),...,(4,9) and you flip the coordinates = 10 pairs

4. Question: How many ways for you to choose a pair from 10 pairs AND a pair from 10 pairs? 10 x 10 = 100
Next: Don't forget, there are 2 ways to arrange 2 chose pairs once we choose one from (y1,y2) and one from (x1,x2) so we have a total of 100 x 2 = 200

The FINAL Answer is: 336 + 200 = 536

Definitely not a GRE question

Thanks for the explanation, indeed time consuming and tough for me.