Let $\phi$ denote Euler's totient function, $p > 1$ be prime, and $n\in\N$. Conjecture and prove a formula for $\phi(p^n)$ in terms of $p$ and $n$.
Question
Let denote Euler's totient function, be prime, and . Conjecture and prove a formula for in terms of and .
Solution
Euler's totient function, denoted as φ(n), is an arithmetic function that counts the positive integers less than or equal to n that are relatively prime to n.
Conjecture: For a prime number p and a natural number n, the formula for φ(p^n) is p^n - p^(n-1).
Proof:
We know that p^n has exactly p^n elements {1, 2, ..., p^n}.
Since p is a prime number, the only divisors of p are 1 and p itself. Therefore, the numbers that are not relatively prime to p^n are the multiples of p, which are {p, 2p, 3p, ..., p^(n-1)*p}.
There are exactly p^(n-1) such numbers because for every number from 1 to p^(n-1), we can multiply it by p to get a number that is not relatively prime to p^n.
Therefore, the count of numbers that are relatively prime to p^n is the total count p^n minus the count of numbers that are not relatively prime to p^n, which is p^(n-1).
So, φ(p^n) = p^n - p^(n-1). This completes the proof.
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