MHB How Can I Solve a Problem Using Euler's Totient Function for Odd Prime Numbers?

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To solve problems using Euler's Totient Function for odd prime numbers, it's essential to understand its definition and application. The discussion begins by examining specific examples, such as $\varphi(5)$, $\varphi(7)$, and $\varphi(11)$, to identify a general rule. For any odd prime number \( n \), the value of \( \varphi(n) \) represents the count of integers less than \( n \) that are coprime to \( n \). This foundational understanding is crucial for tackling more complex problems involving odd primes. Engaging with these examples helps clarify the function's properties and its implications.
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Hello everyone, can anybody help me with this problem?

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The solution is for all odd prime numbers, but I have no idea how to solve it.
Any help will be greatly appreciated.
 
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Hi goody,

Let's take this one step at a time by first looking at the right hand side, then the left hand side for odd primes. We will worry about the non-odd prime case later.

Let's think of some examples first. What is the value of $\varphi(5)$? How about $\varphi(7)$ and $\varphi(11)$? Can you see a general rule emerging from these examples? Now, if $n$ is an odd prime, what should the value of $\varphi(n)$ be? In other words, how many numbers less than $n$ do not share a common divisor with $n$?
 

Thread 'Formal derivation of statement from Peano Arithmetic system'

I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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