## Euler Totient or Phi Function

I have been meaning to write a post on Euler Phi for a while now, but I have been struggling with its proof. I heard it required the Chinese Remainder Theorem, so I have been pushing this until I covered CRT. But recently, I found that CRT is not required and it can be proved much more easily. In fact, the proof is so simple and elegant that after reading it I went ahead and played Minecraft for 5 hours to celebrate.

# Problem

Given an integer $N$, how many numbers less than or equal $N$ are there such that they are coprime to $N$? A number $X$ is coprime to $N$ if $gcd(X,N)=1$.

For example, if $N = 10$, then there are $4$ numbers, namely ${1,3,7,9}$, which are coprime to $10$.

This problem can be solved using Euler Phi Function, $phi()$. Here is the definition from Wiki:

In number theory, Euler’s totient function (or Euler’s phi function), denoted as $\phi(n)$, is an arithmetic function that counts the positive integers less than or equal to n that are relatively prime to n. – Wiki

That’s exactly what we need to find in order to solve the problem above. So, how does Euler Phi work?

# Euler Phi Function

Before we go into its proof, let us first see the end result. Here is the formula using which we can find the value of the $phi()$ function. If we are finding Euler Phi of $N = p_1^{a_1}p_2^{a_2}…p_k^{a_k}$, then:

$$\phi(n) = n \times \frac{p_1-1}{p_1} \times \frac{p_2-1}{p_2}… \times \frac{p_k-1}{p_k}$$

If you want you can skip the proof and just use the formula above to solve problems. That’s what I have been doing all these years. But I highly recommend that you read and try to understand the proof. It’s simple and I am sure someday the proof will help you out in an unexpected way.