Calculate square root with pencil and paper
Introduction
In this article the ancient art of calculating a square root, using pencil and paper, is rediscovered.
The reader should know the addition and multiplication tables by heart .

After the examples an explanation follows why this method is correct.

Notation: the square root of 100 is written as SQRT(100).

Examples
1.
We calculate SQRT(4096)

step 1:
    split the number (rigth to left) in groups of 2 digits, result 40 | 96
step 2:
    find the digit which square is closest (below or equal) to 40,
    this is the 6 , because 6 * 6 = 36
step 3:
    subtract 36 from 40 and shift next 2 digits in place 
	SQRT(40 96)  =  6
	     36
	     ---
	      4 96
The temparary answer is 6. The remainder is 496

Step 4:
    multiply temporary answer by 2, so 2 * 6 = 12
Step 5:
    write 12 as 12? * ?
step 6:
    find digit ? for value closest (below) or equal to remainder.
    This digit is 4 , because 124 * 4 = 496
step 7:
    subtract and add digit 4: 
	W(40 96)  =  64
	  36
	  ---
	   4 96
	   4 96
	   ------
	      0
SQRT(4086) = 64, because 64 *64 = 4096

2.
Calculating SQRT(1522756)

- split number in groups of 2 digits, add 0 when number of digits is odd
    01 | 52 | 27 | 56
- find first square = 1 
	SQRT(01 52 27 56)  =  1
	      1
	     ---
	      0 52
- multiply by 2 : 2 * 1 = 2
- write 2? * ?
- ? = 2
- 22 * 2 = 44
	SQRT(01 52 27 56)  =  12
	      1
	     ---
	      0 52
	        44
	        ---
		   8 27
- multiply temporary answer by 2 : 2 * 12 = 24
- write 24? * ?
- ? = 3, because 243 * 3 = 729 
	SQRT(01 52 27 56)  =  123
	      1
	     ---
	      0 52
	        44
	        ---
		   8 27
		   7 29
		   ------
		     98 56
- multiply temporary answer by 2 : 2 * 123 = 246
- write 246? * ?
- find ? = 4 ( 3 too small, 5 too large)
- 2464 * 4 = 9856 
	SQRT(01 52 27 56)  =  1234
	      1
	     ---
	      0 52
	        44
	        ---
		   8 27
		   7 29
		   ------
		     98 56
		     98 56
		     -----
		         0

3.
Calculate SQRT(5).
Because 5 is no square, the answer will be an approximation.
- write 05| . 00 | 00 | 00 ................
- first square = 2 
	SQRT(05)  =  2.
	      4
	     ---
	      1 00
place decimal point in answer, but do not use in calculations.

- double answer : 2 * 2 = 4
- write 4? * ?
- ? = 2
- 42 * 2 = 84
	SQRT(05)  =  2.2
	      4
	     ---
	      1 00
	        84
	        ---
		   16 00
- double answer : 2 * 22 = 44
- write 44? * ?
- find ? = 3
- 443 * 3 = 1329
- subtract
	SQRT(05)  =  2.23
	      4
	     ---
	      1 00
	        84
	        ---
		   16 00
		   13 29
		   -----
		    2 71 00
- double answer : 2 * 223 = 446
- write as 446? * ?
- ? = 6
- 4466 * 6 = 26796
	SQRT(05)  =  2.236
	      4
	     ---
	      1 00
	        84
	        ---
		   16 00
		   13 29
		   -----
		    2 71 00
		    2 67 96
		    -------
		       3 04
- double answer : 2 * 2236 = 4472
- write as 4472? * ?
- ? = 0, so only shift down 2 zero digits
	SQRT(05)  =  2.2360
	      4
	     ---
	      1 00
	        84
	        ---
		   16 00
		   13 29
		   -----
		    2 71 00
		    2 67 96
		    -------
		       3 04 00
- reapeat steps before for more digits in answer.

Why this works
Core of the method is the product
    (a + b)2 = a2 + 2ab + b2
For clarity this notation is introduced:
if a number consists of digits a and b then
    [ab] = 10a + b
so
    [abc] = 100a + 10b + c
    [23b] = 230 + b
    [(2*3)b] = [6b] = 60 + b
Example : SQRT(4096) again.
Let the answer be [ab] = 10a + b so
    4096 = [ab]2 = (10a + b)2 = 100a2 + 2*10*ab + b2 = 100a2 + (20a + b)b
We have to find digits a and b.
Write number as 40.96
    62 = 36 < 40 and 72 = 49 > 40,
    a = 6
    [6b]2 = 36 00 + 2*10*6b + b2 = 4096
    120b +b2 = (120 + b)*b = [12b]*b
    [12b]*b = 4096 - 36 00 = 496
    b = 4
but this is exactly what we did before, finding the value of ? in 12? * ? .
Only, b is used instead of ?

Another example
SQRT(01 52 27 56)
Write as SQRT( 01 . 52 27 56) and we observe
    a = 1
    write number as 1 52.27 56
    [1b]2 = 1 00 + 2*10*b + b2 = 152 ,27 56
    20b + b2 = (20 + b)*b = [2b]*b
    [2b]*b = 1 52, 27 56 - 100 = 52, 27 56
    b = 2
The temporary answer [12].
Write original number as 152 27. 56
Multiply temporary answer by 10, result is [12b]
So far, number 1 52 27 . 56 - [120]2 is the remainder:
    1 5227, 56 - 14400 = 827, 56
And this is the tric:
regard [12] as the new value of a and find value of b in [12b]

Following instructions above, we find b = 3, so the new temporary answer is [123]
Call [123] the new value of a.
Again shift decimal point 2 places right and multiply temporary answer by 10
    1 52 27 56 - 12 302 = 9856
Calculated value b = 4 and remainder is 0.

The general explanation
Say the root of an 8 digit number xx xx xx xx must be calculated.
    stepaction
    1write number as xx . xx xx xx = a . ------.2
    2calculate a (temporary answer) and subtract : remainder = number - a2
    3multiply remainder by 100, temporary answer by 10
    4calculate b in [ab]
    5calculate new remainder = remainder - [(2a)b]*b
    6new a = [ab]
    7repeat steps 3 .. 7
Do not worry about tehe decimal point : it is automatically right.
Two digits of the number produce one digit of the answer.
The purpose of the decimal point was only to facilitate the explanation.