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Totient Function.md

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+#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:
+
+$$
+\phi(n) = n \prod _{i = 1}^l (1 - \frac 1 {p_i})
+$$