Why can $φ(p_1^{k_1}p_2^{k_2}\cdot \cdot \cdot p_m^{k_m})$ be split into $φ(p_1^{k_1})φ(p_2^{k_2})\cdot \cdot \cdot φ(p_m^{k_m})$?

Conditions

• $n = p_1^{k_1}p_2^{k_2}\cdots p_m^{k_m}$ is the prime factorization of $n$.
• $p_i$ are distinct prime numbers.
• $\varphi$ is multiplicative for coprime integers.

Reasoning, step by step

1. Recall the multiplicativity property of Euler's totient function: if $\gcd(a, b) = 1$, then $\varphi(ab) = \varphi(a)\varphi(b)$.
2. Identify that in the prime factorization $n = p_1^{k_1}p_2^{k_2}\cdots p_m^{k_m}$, the bases $p_i$ are distinct primes.
3. Deduce that powers of distinct primes are coprime, i.e., $\gcd(p_i^{k_i}, p_j^{k_j}) = 1$ for $i \neq j$.
4. Apply the multiplicativity property iteratively to split $\varphi(n)$ into $\varphi(p_1^{k_1})\varphi(p_2^{k_2})\cdots\varphi(p_m^{k_m})$.

Example

Let $n = 12 = 2^2 \cdot 3^1$. The factors $2^2=4$ and $3^1=3$ are coprime because 2 and 3 are distinct primes. Thus, $\varphi(12) = \varphi(4 \cdot 3) = \varphi(4)\varphi(3)$. We know $\varphi(4) = 4 - 2 = 2$ and $\varphi(3) = 3 - 1 = 2$. So $\varphi(12) = 2 \cdot 2 = 4$.

Common misconceptions

• Assuming $\varphi(nm) = \varphi(n)\varphi(m)$ holds for ALL integers $n$ and $m$, not just coprime ones.
• Forgetting that the prime factors in the factorization must be distinct for this direct splitting to work without additional adjustments.

Watch the explanation

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