If this is supposed to be a GMAT question and then I assume that the vertices talked about here are of the square, where one of the vertices must be on origin. Then only 4 squares can be drawn.

What is the answer? Is it A

Will explain if its correct.

A is not the answer niks and it is a GMAT question.

I will give the answer after a few more responses.

My guess is E. You can rotate the square along the vertex at the origin and 360 degrees. The 4 obvious positions are where two sides align with the x and y axis. However, there should be two other positions in each quadrant where the vertex will just so happen to fall on the integer coordinates. What's the OA?

I will go with 12.

One vertex can lie on origin and one on any of these points

(10,0), (-10,0), (0,10), (0,10), (6,8), (6,-8), (-6,8), (-6,-8), (8,6), (8,-6), (-8,6), (-8, -6)

From any of these points and origin you can calculate other 2 points.

e.g. Suppose 3rd coordinate is at (x,y) - diagonally opposite to origin

so, distance between (x,y) and (6,8) will be 10

and distance between (x,y) and origing will be 10root2

For this example other 2 points will be (2,14) and (-3,4)

I don't think you would really need to calculate for each of these points individually. If you are able to prove one point fits the formala, I think you can safely assume others will satisfy too.

I would definitely like to see some other answers.

E is the answer.

Kajcha, I think you explanation is spot on. Kudos.

I agree with Kajcha the ans should be E 12, I solved it like
Kajcha only.

Agree with 12. Thanks, kajcha for a good explanation.