Euler's totient function gives all the numbers that are relatively prime to $n$, that are below $n$. This is Euler's formula:
$$\psi(n) = n \prod_{p|n} \left(1 -\frac{1}{p}\right) $$
In order to use the formula, we must first prime factorize $n$. Let's take a smaller number first. Suppose you want to use $\psi(n)$ for $36$ as stated in the wiki article. Then you would take the following steps.
- Prime factorize $n$
- Then apply "sub" (distinct) $p$ in and continue to do so until you have run out of distinct $p$'s
So, first for the prime factorization of $36$ or any number. Here's the process.
Divide n by p
if n mod p = 0 then continue to divide by n
if not then move onto the next prime
You are done when you are left with a prime number
For $36$ it would be the following process:
$36/2 = 18 \rightarrow $18 is still divisible by 2, so continue
$18/2 = 9 \rightarrow 9$ is odd and therefore not divisible by $2$. Move onto the next prime, $3$
$9/3 = 3 \rightarrow 3$ is a prime and you are complete.
Therefore, the prime factors of $36$ are: $2,2,3,3$ or in other words $2^2*3^2$. Once we have the prime factors of $n$ we can use the function.
$$\begin {align}
&\psi(n) = n \prod_{p|n} \left(1- \frac{1}{p}\right)\\
&\psi(36) = 36 \prod_{p|n} \left(1- \frac{1}{2}\right)\left(1- \frac{1}{3}\right)\\
&=12
\end{align}$$
Therefore, there are $12$ below $36$ that are relatively prime to $36$.
For the example that you have provided it would look something like this:
$$\begin {align}
&\psi(n) = n \prod_{p|n} \left(1- \frac{1}{p}\right)\\
&\psi(93296) = 93296 \prod_{p|n} \left(1- \frac{1}{2}\right)\left(1- \frac{1}{7}\right)\left(1- \frac{1}{17}\right)\\
&=37632
\end{align}$$
This is the basic process to use the Totient function, note that there are various other formulas that one can use, I find that this one is the easiest to understand. Furthermore, prime factorization should not be difficult if you know the primes below $100$.
P.S For practice, you can verify your answers here
$36=4\times9$
Since $4$ and $9$ are relatively prime, we have
$\phi(36)=\phi(4)\phi(9)$
Using the identity that for any prime $p$, $\phi(p^n)=p^n-p^{n-1}$
$\phi(4)\phi(9)=\phi(2^2)\phi(3^2)=(2^2-2).(3^2-3)=12$
As for the second part of the question, then, 13788 (mod 12) (mod 36) $\ne
138(mod 36)$ and hence your question is wrong.
There seems to be typo error and this part of the question can be modified to $13^{788}(mod\ 36) = 13^{788(mod \phi(36))}(mod\ 36) = 13^{788(mod 12)}(mod\ 36)=13^8(mod\ 36)=25$
Best Answer
Well, @Arthur cleared this up for me in the comments, so I'll answer my own question:
$\phi(ab)$ = $\phi(a)$ $\times$ $\phi(b)$, only if a and b are co-prime.
So, while $\phi(100)$ = $\phi(25)$ $\times$ $\phi(4)$ because 25 and 4 are co-primes, $\phi(100)$ = $\phi(5)$ $\times$ $\phi(5)$ $\times$ $\phi(2)$ $\times$ $\phi(2)$ is not true because the 2s are not coprime, and the 5s are not co-prime either.
So, $\phi(100)$ = $\phi(25)$ $\times$ $\phi(4)$.
$\phi(25)$ = 20 (We can evaluate this through the formula $\phi(p^n) = p^{n-1}(p-1)$, so $\phi(5^2) = 5^{1}(4) = 5 \times 4 = 20.
$\phi(4)$ = 2.
$\implies$$\phi(100)$ = $20 \times 2$ = 40.
Thanks to @Arthur and @DreiCleaner for clearing this up, and @J.W.Tanner for suggesting some ways to make this answer better!