How can I find the square root using pen and paper? Okay, I know this is very basic question. I learned 2 methods in school. But now, I forget one. 
Here is a simple method that I know.


*

*Find the prime divisors of the number

*Omit the half of numbers that have been appeared even times

*multiply the rest


For example you want to find square root of 36. You find the divisors. They are 2x2x3x3. In step 2 they appeared as 2x3. That is 6. Problem is this method works when the square root is an integer number. It doesn't work for numbers that doesn't have integer square root. Like 38.
So my question is how can I find the square root of any arbitrary number using pen and paper?
 A: If your number is close enough to a perfect square, you can use the expansion
$$
\sqrt{1+x} = 1 + \frac{1}{2}x - \frac{1}{8}x^2 +- \cdots + (-1)^{n-1}\frac{(2n-3)!!}{2\cdot(2n)!!}x^n+\cdots
$$
where
$$
(2n)!! = (2n)(2n-2)\cdots4\cdot2,\quad(2n+1)!!=(2n+1)(2n-1)\cdots3\cdot1.
$$
A famous physist R. Feynman is said to have used this formula and beat an abacus. 
A: This comes from Euler's method in approximating solutions to differential equations.  This begins with the fundamental notion that the derivative of f(x)=$x^.5$ is equal to f'(x)=$.5x^{-.5}$.  So now it is possible to approximate roots.  Take the root of 38.  The nearest perfect square is 36.  so to approximate the square root of 38, you want to take the square root of 36 and then add $2*.5x^{-.5}$.  This gives you $6+1/6$ or 6.166666 repeating, which is a pretty good approximation.  If you have more specific questions, just ask.
A: The general form of Newton's Method for zero-finding is 
\begin{equation*}
x_{k+1} := x_k - \frac{f(x_k)}{f'(x_k)}
\end{equation*}
 The Reciprocal Square Root Newton algorithm is obtained by setting $f(x) = \frac{1}{x^2-a}$ and we get
\begin{equation*}
x_{k+1} := x_k - \frac{f(x_k)}{f'(x_k)} = x_k - \frac{1/x_k^2-a}{-2/x_k^3}=x_k +
\frac{1}{2}x_k(1-ax^2_k).
\end{equation*}
The sequence $\{x_0,x_1, \ldots, x_k, \ldots\}$ converges quadratically, for all $x_0 >0$, to a fixed point $x$. That is
\begin{equation*}
x = x + \frac{1}{2}x(1-ax^2), 
\end{equation*}
which has two solutions, $x = 0$, and $x = 1/\sqrt{a}$.
We compute $x_{k+1}$ in two steps:


*

*$e_k := 1-a\star x_k\star x_k $

*$x_{k+1} := x_k + e_k\star x_k/2$


Here,  $e_k$ is the relative error in $x^2_k,$ i.e.,
$(1/a-x^2_k)/(1/a)=(1-ax^2_k),$ and  $e_kx_k/2$ is the correction term. 
The most important aspect of this algorithm is that no divisions are
required. The division by 2 can be done by bit-shifting. Calculating $x_{k+1}$ requires 3 Mults and 2 Adds. One final multiplication, $a\star x_k$, is done to calculate $\sqrt{a}$.
If a suitable starting value for $x_0$ is chosen, then this algorithm converges to full IEEE double precision in about 5 iterations. An algorithm similar to this is used by Intel in their Itanium processor, whose basic arithmetic operation is the fused multiply-add (FMA) instruction, $z = ax+y$. All other operations are performed using this FMA instruction. For historical reasons, Intel shies away from long division.
See Peter Markstein, IA-64 and Elementary Functions, Prentice-Hall, 2000
Additional Information
The Reciprocal Newton algorithm is obtained by setting $f(x) = \frac{1}{x-a}$ and we get
\begin{equation*}
x_{k+1} := x_k - \frac{f(x_k)}{f'(x_k)} = x_k - \frac{1/x_k-a}{-1/x_k^2}=x_k +
x_k(1-ax_k).
\end{equation*}
We compute $x_{k+1}$ in two steps:


*

*$e_k := 1-a\star x_k $

*$x_{k+1} := x_k + e_k\star x_k/2$


Here,  $e_k$ is the relative error in $x^2_k,$ i.e.,
$(1/a-x_k)/(1/a)=(1-ax_k),$ and  $e_kx_k/2$ is the correction term. Each of these two steps is calculated by a single FMA instruction. The division, $z = y/x$, is calculated as $y\star\frac{1}{x}$. Note how similar this algorithm is to the reciprocal square root algorithm.
Choosing a starting value $x_0$ for the Reciprocal Algorithm is a delicate matter. It can be shown that the domain of convergence is $ 0 < x_0 < \frac{2}{a}$, but this begs the question: what is $\frac{2}{a}$? This is getting into deep water, so I'll leave the answer to Peter Markstein above.
A: The best method I know of is the recursive series:$$x_1=b\;\;\;\;\;\;\;\;\;x_{n+1}=\frac{1}{2}\left(x_n+\frac{b}{x_n}\right)$$  It converges very rapidly to $\sqrt{b}$ - for example for b = 3, it is accurate to 7 decimal places after only 4 terms.  The long division might not be very easy to carry out strictly with pencil and paper, but it is doable.  
A: Just found an interesting way to find square root of a number using Derivatives. All the above answers are awesome, but this method just blew my mind.This method  is a trade off between accuracy and time taken to get the result.
Lets take an example. Say $\sqrt{53}$.
Consider $y = \sqrt{x}$ graph

We have $$y = \sqrt{x}$$
$$dy = \frac{1}{2\sqrt{x}}dx$$
Now, at $x\approx49$ and $dx=4$
$$dy = \frac{1}{2\sqrt{49}}dx$$
$$dy = \frac{2}{7} \approx 0.28$$
Therefore $$\sqrt{53} = 7 + dy = 7.28$$
The results might get a little inaccurate after $\approx$100 because the perfect squares get more sparsely populated and thus making $dx$ comparatively nearer to $x$. Therefore always try to  take out all the integers you can from inside the square root and then apply this method on whatever remains inside.
Obviously for this, you need to know the nearest (just less than) perfect square to the given number. But I guess, a little trial and error should get you that perfect square.
