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Prove that $\phi(xy) = \phi(x) \phi(y)$ for any $x$ and $y$ with $(x, y) = 1$.

I understand the concept, and have done several examples proofing this but cannot put it in "proof form" because unless its a counterexample, it cannot be proven by just using the numbers that I have from a table.

Martin Sleziak
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T.Turn
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    Perhaps you should mention what $\varphi$ is. Presumably you're talking about Euler's totient, but there is nothing to indicate this is the case. – Hayden Jul 11 '14 at 03:48
  • Yes I am talking about Euler's totient – T.Turn Jul 11 '14 at 03:50

2 Answers2

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The Chinese remainder theorem indicates we can always separately solve

$$\begin{cases}a\equiv b_1\mod p_1^{\alpha_1} \\ \vdots \\a\equiv b_n\mod p_n^{\alpha_n}\end{cases}$$

for distinct primes $p_i$

So since there are $\varphi(p_i^{\alpha_i})$ choices for equivalence classes mod $p_i^{\alpha_i}$ coprime to the modulus for each of the independent choices, the total number of numbers coprime to $n$ is

$$\varphi(n)=\prod_{i=1}^n\varphi(p_i^{\alpha_i})$$

Adam Hughes
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If you've seen a proof of the formula $$\sum\limits_{d \mid n} \phi(d) = n$$ ($d$ runs through all the positive divisors of $n$, including $1$ and $n$), you can actually deduce the multiplicativity of $\phi$ from here. The above rule itself is not hard to prove, and requires little more than good intuition about the greatest common divisor of two numbers. I can give you a proof of the above rule if you want.

For example, let's use the above formula to show that $\phi(pq) = \phi(p)\phi(q)$ for distinct primes $p$ and $q$. The divisors of $pq$ are $1, p, q, pq$ so $$pq = \phi(1) + \phi(p) + \phi(q) + \phi(pq) = 1 + (p-1) + (q-1) + \phi(pq)$$ which implies $$\phi(pq) = pq + 1 - p - q = (p - 1)(q-1) = \phi(p) \phi(q)$$ The general case of multiplicativity can be similarly done, although it requires some tricky induction.

D_S
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