Having the number 59 for example find $x$ which $x^2$ is closer but lower to the 59. In this case is 7 I don't know how to name this but this is what I would need.
42  43  44  45  46  47  48  49
41  20  21  22  23  24  25  50
40  19  6   7   8   9   26  51
39  18  5   0   1   10  27  52
38  17  4   3   2   11  28  
37  16  15  14  13  12  29  
36  35  34  33  32  31  30  

This is a map of blocks. It starts by 0 and the go 1, 2, 3, 4, 5.... circular clockwise. So given that I know the number 52, in this case I would like to know it's $(x, y)$, which in this case is $(x=4,\ y=0)$.
The first thing I need is:


*

*Having the number 59 for example find $x$ which $x^2$ is closer but lower to the 59. In this case is 7.


Which gives me the magnitude, to know how away from 0 is, then based on that I have the length of the square, this means how many blocks are on each side. In the case of 52, I get the number 7 based on that: $52-49+1 = 4$. With the 4 then I will make a loop until it finds the position. It will turn after 8 times.
Any help is appreciated, thanks very much!

You work by taking it apart.  I will use $n$ for the number you are looking for the position of, as you use $x,y$ as the coordinates.  As you say, you first want to find $m=\lfloor \sqrt n \rfloor$, the greatest integer less than or equal to the square root of $n$, as $m^2$ is the last upper right or lower left corner before $n$   
If $m$ is even, its coordinates are $(-\frac m2,-\frac m2)$.   Now you need to count $n-m^2$ spaces beyond.  If $n \le m^2+m+1$, it is in the column just to the left of $m^2$ and its coordinates are $(-\frac m2-1,-\frac m2+n-m^2-1)$...
If $m$ is odd, its coordinates are $(\frac {m+1}2,\frac {m+1}2-1)$...
The ...s have more reasoning like the second sentence of the second paragraph.  I leave that as an exercise.
