Totient multiplicativity proof for primes

$\phi(xy) = \phi(x)\phi(y)$

Definition

$\phi(n)$ is defined as the function which calculates the number of coprimes of $n$ which are $< n$, that is, every number $< n$ which doesn't share cofactors with $n$. We could have the exact same definition by implying that $\phi(n)$ is the number of coprimes of $n$ which are $\leq n$, and since $n$ is never coprime with itself it follows trivially that we are computing the number of coprimes of $n$ which are $< n$.

Example

Let's start with an example with $2$ simple primes. It looks stupid but it's easily formalizable and works for every pair of primes.

$10 = 2 \cdot 5$
$\phi(5) = |\{ 1, 2, 3, 4 \}| = 4$
$\phi(2) = |\{ 1 \}| = 1$
$->$
$\phi(10) = |\{1, 3, 7, 9 \}| = \phi(2)\phi(5) = 1 \cdot 4 = 4$

• $5$ numbers share the $2$ divisor with $10$
• $2$ numbers share the $5$ divisor with $10$

From these we have to remove $1$ element because $10$ is considered twice $(2(5), 5(2))$ then

$5 \cdot 2 - (5 + 2 - 1) = 4$
$->$
$\phi(10) = 4$

This is easily formalizable because for all pair of primes $m, n$, their multiplication will be the only element considered twice. It's also worth noting that the previous removal can be easily considered an addition, that is, if we follow the previous picture structure, we can see that $10$ will be removed twice if we consider the removal of the rightmost column and lowest line (which was my initial intuition). Indeed

$5 \cdot 2 - (5 + 2 - 1) = 5 \cdot 2 - 5 - 2 + 1$

Formalization

We take 2 primes $m,n$ then

$\phi(mn) = m \cdot n - (m + n - 1)$
$->$
$m \cdot n - m - n + 1 = (m - 1)(n - 1)$

where

$\phi(m) = m - 1$
$\phi(n) = n - 1$

Further refinition

The scheme above looks a little bit messy, and seems like it does not follow a structured rule, let's refine it taking another example.

Example 2: $\phi(3 \cdot 7)$

Before delving the next example, let's refine the scheme above to make it more structured and understand my point.

We can see that $[2(2) + 1 = 5]$ is not positioned in the last line but we could simply swap that element with $[2(4)+1]$ and the game would be done.

Considering $3 \cdot 7$ we know that there will be $3$ elements which share $7$ as co-factor and so even if the structured form proposed doesn't place them in the last line it will be always possible to swap them.

Taken $a, b$ where $b > a$ there will be $a$ elements which share the $b$ cofactor, this means that they will always be positionable in the last line (of $a$ elements).