1.9 KiB
#Math #NT
Definition
Euler’s totient function returns the number of integers from 1 \leq k \leq n for a positive integer n. It is notated as:
\phi(n)
\phi(n) for Prime Powers
Through prime factorization, for p^k, the only positive integers below p^k where \gcd(p^k, n) > 1 is where n = mp, for 1 \leq m \leq p^{k - 1}. Therefore:
\phi(p^k) \\ = p^k - p^{k - 1} \\ = p^{k - 1}(p - 1) \\ p^k(1 - \frac 1 p)
Multiplicative Property of \phi
If m and n are coprime:
\phi(m)\phi(n) = \phi(mn)
Proof: Let set A be all numbers coprime to m below m, and set B be all numbers coprime to n below n.
|A| = \phi(m) \\ |B| = \phi(n)
Let set D be all possible ordered pairs using elements from A and B, where the element of A is first. If for each element $(k_1, k_2)$in set D we return a value \theta where:
\theta \equiv k_1 \mod m \\ \theta \equiv k_2 \mod n
CRT ensures \theta is unique to \mod ab and exists. Given the fact \gcd(x + yz, z) = \gcd(x, z), we can say that:
\gcd(\theta, m) = \gcd(k_1, m) = 1 \\ \gcd(\theta, n) = \gcd(k_2, n) = 1 \\ \gcd(\theta, mn) = 1
If we put all \theta in set C, we can see that set C has all the elements fitting the above conditions. Looking at the length of C:
|C| = \phi(mn) \\
|C| = |A| * |B| = \phi(m)\phi(n) \\
\phi(mn) = \phi(m)\phi(n)
Value of \phi for any Number
Let a positive integer n prime factorization be:
n = p_1^{k_1}p_2^{k_2}p_3^{k_3}...p_l^{k_l}
Now using the properties above:
\phi(n) \\
= \prod _{i = 1}^l \phi(p_i^{k_i}) \\
= \prod _{i = 1}^l p_i^{k_i}(1 - \frac 1 {p_i})
Multiplying all p_i^{k_i} gives n, so factor that out:
= n \prod _{i = 1}^l (1 - \frac 1 {p_i})
(you can derive most textbook definitions from this formula easily)
Final formula:
\phi(n) = n \prod _{i = 1}^l (1 - \frac 1 {p_i})