Proving that $\varphi(n)=n\prod (1-1/p)$ without using multiplicativity

Question

$$\varphi(n)=n\prod_{p \ \text{prime}} (1-1/p)$$

Can this useful formula be derived without using the fact that Euler's totient function is multiplicative?

Answer 1

Let $n=\prod_{i=1}^{k}p_{i}^{a_i}$ where the $p_i$'s are distinct primes and the $a_i$'s are all $\geq1$. Let $A_i$ denote the subset of $\{1,2,\ldots, n\}$, all of whose elements are divisible by $p_i$. Then the set $A := \{m|1 \leq m \leq n, {\rm gcd}(m,n) \gt 1\}$ is precisely the union of the $A_i$'s and we have, from the principle of inclusion-exclusion,

$$|A|=\sum_i|A_i| - \sum_{i<j}|A_i \cap A_j| + \sum_{i<j<k}|A_i\cap A_j \cap A_k|-\cdots$$

which gives us,

$$|A|=\sum_i \frac{n}{p_i}-\sum_{i<j}\frac{n}{p_ip_j}+\sum_{i< j < k} \frac{n}{p_ip_jp_k}-\cdots.$$

Thus, $$\phi(n)=n-|A|=n-\sum_i \frac{n}{p_i}+\sum_{i<j}\frac{n}{p_ip_j}-\sum_{i< j < k} \frac{n}{p_ip_jp_k}+\cdots$$ giving us the result, $$\phi(n)=n\prod_{i=1}^{k}(1-\frac{1}{p_i})=n\prod_{p|n}(1-\frac{1}{p}).$$

Answer 2

Well, you need some property. Pick a random number from $1$ to $n$. Given prime divisor $p$ of $n$, what is the probability that your random choice is not divisible by $p$? It is $1-\frac1p$. Then you need to prove that these probabilities are independent - that is, if we pick a random number from $1$ to $n$ not divisible by $p_1,\dots,p_k$, then the (conditional) probability that it is not divisible by some different $p$ is still $1-\frac1p$.

This ultimately shows that the probability that a random number from $1$ to $n$ is relatively prime to $n$ is:

$$\frac{\phi(n)}{n} = \prod_{p\mid n} \left(1-\frac1p\right)$$

Proving that independence is essentially the same as proving multiplicative, but it has a different flavor.

Another method is to do an inclusion-exclusion argument.

Answer 3

The theorem itself is partially hinting on the multiplicative property of $\varphi$ since it is asking you to separate in the primes.

I don't know if the following argument is useful.

Let $n=p_1^{\alpha_1}p_2^{\alpha_2}\dots p_r^{\alpha_r}$ then a number $a$ is coprime to $n$ if and only if $a$ is congruent to a number that is not a multiple of $p_i\bmod p_i^{\alpha_i}$ for each $i$. Since the set $\{1,2,3\dots n\}$ is a complete residue system and by the chinese remainder theorem there is exactly one solution $\bmod n$ to each system of the form:

$x\equiv a_1 \bmod p_1^{\alpha_1}$

$x\equiv a_2 \bmod p_2^{\alpha_2}$

....

$x\equiv a_r \bmod p_r^{\alpha_r}$

we conclude the number of number relatively prime to $n$ in the set $\{1,2,3\dots n\}$ is equal to the number of ways to place suitable $a_i$'s to the conguence system above. Of course there are $p^{\alpha_i-1}(p-1)$ ways to do so for each $a_i$. Therefore $\varphi(n)=\prod\limits_{i=1}^rp^{\alpha_i-1}(p-1)=n\prod\limits_{i=1}^r\frac{p-1}{p}$

Answer 4

Yes this can be shown via inclusion-exclusion but is a bit cumbersome.

I will spell it out up to three distinct prime-divisors only.

First recall that $1\le a \le n$ is coprime to $n$ when it is not divisible by any prime $p$ that divides $n$.

Let $p \mid n$. The number of $1\le a \le n$ such that $p \mid a$ is $n/p$ so the number not divisible by $p$ is $n - n/p = n(1 - 1/p)$. So the number of $1\le a \le n$ not divisible by $p$ is $ n(1 - 1/p)$ which is $\varphi(n)$ if $p$ is the only prime dividing $n$.

If we have two distinct primes $p,q$ dividing $n$ and denote the set of multiples $1 \le a \le n$ of $p$ by $A_p$ and likewise define $A_q$ then we have $|A_p| = n/p$ and $|A_q| = n/q$ and $|A_p \cup A_q| = |A_p| + |A_q| - |A_p \cap A_q|$. Now $A_p \cap A_q$ is the set of all $a$ divisible by $p$ and $q$ so its cardinality is $n/(pq)$.

Thus $n - |A_p \cup A_q| = n - n/p - n/q + n/(pq)= n (1-1/p)(1-1/q)$. This is the number of $1 \le a \le n$ not divisible by $p$ or $q$, so it is $\varphi (n)$ in case $n$ is divisible only by the primes $p$ and $q$.

If we now have three distinct primes $p,q,r$ we get in a similar way

$|A_p \cup A_q \cup A_r| = |A_p | + |A_q|+ |A_r|- |A_p \cap A_q| - |A_p \cap A_r| - |A_r \cap A_q| + |A_p \cap A_q \cap A_r|$.

So $$n - |A_p \cup A_q \cup A_r|= n - n/p - n/q -n/r + n/(pq) + n/(pr)+ n/(qr)- n/(pqr)= n(1-1/p)(1-1/q)(1-1/r).$$ And this is $\varphi(n)$ when $n$ is divisible only by $3$ primes.