Euler Phi Function is Multiplicative

Theorem
The Euler $\phi$ function is a multiplicative function:


 * $$m \perp n \Longrightarrow \phi \left({m n}\right) = \phi \left({m}\right) \phi \left({n}\right)$$

where $$m, n \in \Z^*_+$$.

Proof
Let $$R = \left\{{r_1, r_2, \ldots, r_{\phi \left({m}\right)}}\right\}$$ and $$S = \left\{{s_1, s_2, \ldots, s_{\phi \left({n}\right)}}\right\}$$ be the reduced residue systems for the respective moduli $$m$$ and $$n$$.

We are to show that the set of $$\phi \left({m}\right) \phi \left({n}\right)$$ integers:
 * $$T = \left\{{n r + m s: r \in R, s \in S}\right\}$$

is a reduced residue system for modulus $$m n$$.

We need to establish the following:
 * Each integer in $$T$$ is prime to $$m n$$;
 * No two integers in $$T$$ is congruent modulo $m n$;
 * Each integer prime to $$m n$$ is congruent modulo $m n$ to one of these integers in $$T$$.

We prove each in turn:


 * Suppose $$p$$ is a prime divisor of $$\gcd \left\{{n r + m s, m n}\right\}$$ where $$r \in R, s \in S$$.

As $$p$$ divides $$m n$$ but $$m \perp n$$, $$p$$ either divides $$m$$ or $$n$$ but not both, from Divisors of Product of Coprime Integers.

Suppose WLOG that $$p \backslash m$$.

Then as $$p \backslash n r + m s$$, we have $$p \backslash n r$$ and hence $$p \backslash r$$.

But then $$p \backslash \gcd \left\{{m, r}\right\} = 1$$ which is a contradiction.

Similarly if $$p \backslash n$$.

So there is no such prime and hence $$n r + m s \perp m n$$.


 * Suppose $$n r + m s = n r' + m s' \left({\bmod\, m n}\right)$$, where $$r, r' \in R, s, s' \in S$$.

Then:
 * $$n \left({r - r'}\right) + m \left({s - s'}\right) = k \left({m n}\right)$$ for some $$k \in \Z$$.

As $$m$$ divides two of these terms it must divide the third, so $$m \backslash n \left({r - r'}\right)$$.

Now $$m \perp n$$ so by Euclid's Lemma $$m \backslash \left({r - r'}\right)$$, or $$r \equiv r' \left({\bmod\, m}\right)$$.

But $$r$$ and $$r'$$ are part of the same reduced residue system modulo $$m$$, so $$r = r'$$.

Similarly for $$n$$: we get $$s = s'$$.

Hence distinct elements of $$T$$ can not be congruent modulo $m n$.


 * Let $$k \in \Z: k \perp m n$$.

Since $$m \perp n$$, from Bézout's Identity we can write $$k = n r' + m s'$$ for some $$r', s' \in \Z$$.

Suppose there exists some prime number $$p$$ such that $$p \backslash m$$ and $$p \backslash r$$.

Such a prime would be a common divisor of both $$k$$ and $$m n$$, contradicting $$k \perp m n$$.

Hence $$r' \perp m$$ and so is congruent modulo $m n$ to one of these integers in $$R$$.

By the same argument, $$s' \perp n$$ and so is congruent modulo $m n$ to one of these integers in $$S$$.

Writing $$r' = r + a m, s' = s + b n$$ we have:
 * $$k = n r' + m s' = n r + m s + m n \left({a + b}\right) \equiv n r + m s \left({\bmod\, m n}\right)$$.

Hence the result.