# Find the square root of an integer with trigonometry and Lagrange's four-square theorem

My friend was solving the following problem during a interview for Citigroup's IT department.

Find the square root of a integer n, without using the built in sqrt function. (The range of the result was not specified, I assume it's double)

This is a common interview question.

There are many ways to do it. I want to come up with a way no one else would think of, something that could amaze the interviewer. I mean, she might interviewed enough people to get bored with the standard answers.

I present the following highly inefficient but somewhat creative solution. The code is here.

How does it work?

We know $n$ is an integer. By Lagrange's four-square theorem, $n=a^2+b^2+c^2+d^2$ for integer $a,b,c,d$. $\sqrt{n} = \sqrt{a^2+b^2+c^2+d^2}$. Thus $\sqrt{n}$ is the magnitude of the vector $[a,b,c,d]$. $a,b,c,d$ can be calculated by brute force search(therefore runs in $O(n^2)$ time).

Note a simple improvement of the naive algorithm can reduce the computation time to $O(n^\frac{3}{2} \log n)$ by doing a binary search for the last square.

A much smarter randomized algorithm by Michael O. Rabin and Jeffrey Shallit have a running time of $O(\log^2 n)$.

A recursive algorithm using the following relation can find the magnitude of any vector(assume $a_i\neq 0$) $|[a_0,...,a_{n-1},a_n]| = \frac{a_n}{\sin(\tan^{-1}(\frac{a_n}{|[a_0,...,a_{n-1}]|}))}$ It's easy to see, this breaks a $n$-dimension vector into orthogonal vectors of $n-1$-dimensions and $1$-dimension. We get a right triangle. Trigonometry comes in handy.