Why does a prime number p satisfy $φ(p) = p - 1$?

Conditions

• $p$ is a prime number.
• $\varphi(p)$ is defined as the count of integers from 1 to $p$ that are coprime to $p$.

Reasoning, step by step

1. Let $p$ be a prime number.
2. Consider the integers from 1 to $p$.
3. Note that all integers from 1 to $p-1$ do not contain $p$ as a factor.
4. Since $p$ is prime, its only factors are 1 and $p$. Therefore, any integer less than $p$ shares no common factors with $p$ other than 1.
5. This means $\gcd(a, p) = 1$ for all $a \in \{1, 2, \ldots, p-1\}$.
6. Count the number of such integers, which is $p - 1$.
7. Conclude that $\varphi(p) = p - 1$.

Example

Let $p=5$. The integers from 1 to 5 are 1, 2, 3, 4, 5. The greatest common divisors with 5 are: $\gcd(1,5)=1$, $\gcd(2,5)=1$, $\gcd(3,5)=1$, $\gcd(4,5)=1$, $\gcd(5,5)=5$. The integers coprime to 5 are 1, 2, 3, and 4. There are $5-1=4$ such integers. Thus, $\varphi(5) = 4$.

Common misconceptions

• Thinking that 1 is not coprime to $p$ (1 is coprime to every positive integer).
• Believing that $\varphi(p)$ counts the number of primes less than $p$ (it counts all integers coprime to $p$, not just primes).

Watch the explanation

Explore next

• How is $φ(p^k) = p^k - p^{k-1}$ derived?
• What is the general formula for Euler's totient function?
• How do you evaluate the definite integral of sqrt(x) from 2 to 3?