Euler Phi Function is Multiplicative

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Theorem

$m \perp n \implies \map \phi {m n} = \map \phi m \map \phi n$

where $m, n \in \Z_{>0}$.

Proof

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

We are to show that the set of $\map \phi m \map \phi n$ integers:

$T = \set {n r + m s: r \in R, s \in S}$

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:

Aiming for a contradiction, suppose $p$ is a prime divisor of $\gcd \set {n r + m s, m n}$ where $r \in R, s \in S$.

We have that $p$ divides $m n$ but $m \perp n$.

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

Without loss of generality, suppose $p \divides m$.

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

But then $p \divides \gcd \set {m, r} = 1$ which is a contradiction.

Similarly if $p \divides n$.

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

$\Box$

Let $n r + m s = n r' + m s' \pmod {m n}$, where $r, r' \in R, s, s' \in S$.

Then:

$n \paren {r - r'} + m \paren {s - s'} = k \paren {m n}$ for some $k \in \Z$.

As $m$ divides two of these terms it must divide the third, so:

$m \divides n \paren {r - r'}$

Now $m \perp n$ so by Euclid's Lemma:

$m \divides \paren {r - r'}$

or:

$r \equiv r' \pmod m$

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 $m n$.

$\Box$

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

Since $m \perp n$, from Set of Integer Combinations equals Set of Multiples of GCD we can write $k = n r' + m s'$ for some $r', s' \in \Z$.

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

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 $m$ to one of these integers in $R$.

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

Writing $r' = r + a m, \, s' = s + b n$ we have:

$k = n r' + m s' = n r + m s + m n \paren {a + b} \equiv n r + m s \pmod {m n}$

Hence the result.

$\blacksquare$