Using the Euler totient function for a large number So I have a test in a couple of hours and I'm having trouble finding information on how to use the Euler totient function for a large number so I'm wondering if someone could give me step-by-step instructions? :)
Here's a sample question for the test:
Find $\psi(93296)$.
Could you also give instructions on finding the prime factorisation as well? I used an online calculator and it is $2^4 * 7^3 * 17$ but I wouldn't know how to calculate this. 
 A: 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
A: $ψ(93296)=ψ(2^4)\timesψ(7^3)\timesψ(17)$
        =$(2^4-2^3)\times(7^3-7^2)\times(17-1)$
        =$8\times294\times16$
        =$37632$
