## 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.

# Problem

Given a number N, determine if it is a prime.

# What is a Prime Number?

A prime number is a positive number greater than 1 which has no positive divisor except for 1 and itself. For example, 2, 3, 5, 11 are prime numbers. Whereas 4, 6 are non-prime or composite numbers.

# O(N) Solution

From the definition, we can easily construct a Primality Test function that can determine if a given number N is prime or not. Let us call it isPrime() function.

bool isPrime ( int n ) {
if ( n <= 1 ) return false; // n needs to be greater than 1
for ( int i = 2; i < n; i++ ) {
// If it is possible to divide n with a number other than 1 and n, then it is not prime
if ( n % i == 0 ) return false;
}
return true; // Otherwise, this is prime
}


The code simply iterates over all values between 2 and N-1 and checks if it can divide N. It has a complexity of O(N).