1. ## cartesian plane....

Hello...

I am not sure if this is the right section to put this in...but I have difficult solving this problem and I would appreciate any help you could provide!

--->In the cartesian place, consider the set of points A = [ (i,j) / 0 <=i, j<=40, with i and j are integers]. How many squares can be formed so that all the corners of a ll squares belong to A, with sides parallel to the x and y axes...

thanks for the help!

2. It's actually easier to do this algebraically, using n rather than 40. So suppose that the vertices are at the points (i,j) with 0 ≤ i,j ≤ n. Now count the number of possible squares according to the length of their sides.

For squares of side 1, the lower left corner of the square can be at any point (i,j) with 0 ≤ i,j ≤ n–1. So there are $n^2$ such squares.

For squares of side 2, the lower left corner of the square can be at any point (i,j) with 0 ≤ i,j ≤ n–2. So there are $(n-1)^2$ such squares.

...

For squares of side n, the lower left corner of the square can can only be at the point (0,0). So there is only 1 such square.

Now you take it from there ... .

[The big advantage of using n rather than 40 is that you can check whether your answer is likely to be correct, by seeing what it tells you for some small value of n, like n=3 say. In that case, you can actually count the number of possible squares and check whether that agrees with the answer given by your formula.]

3. thank you very much Sir!

but does that mean...since there are 4 corners of the squares...so there are 4 times the number of squares we found for the lower left corner of the square?

which is 4...for n sides of the square?
I am a little confused....

thanks again Sir!