Recent content by xWhiteyx

thanks. I'll probably be getting back to you later on harder phi functions. This is still the easy bit.

Could you then say that for P^3, it's basically the same thing, except excahnge P^n for P^3.

Sorry, I'll try to be more clear. I'll show you what I mean by giving you my answer from an earlier quiestion question. 1) when p is a prime number, ontain an expression for phi(p) in terms of p:- Numbers which arn't coprime to p would be divisible by p. All intergers from 1 to p-1 are not...

phi(3^4)=54. 81-54=27 which is 3^3. So if I assume, before I find it out, that phi(3^5) would be 243-81 (162) and then figure it out. The answer is indeed 162. So would phi(P^n)=(P^n)-(p^n-1)? If so, how would you right that as an expression. My teacher said that it is all well and good...

9. So, the number of numbers coprime to, for example, 3^3 is 3^3- 3*3. Would that then form the formula: phi(P^n)=(P^n)-(P*P) If that is the formula, then how would you express that?

But it asks us to obtain an expression in terms of p, without using additional numbers. If I could do that, then this homework would be easy but it isn't. So far, i believe that this might get me somewhere: Take 3^3 for example. Which numbers less than 27 share a factor with it? Clearly 1...

This is getting on my nerves now. I am stumped on these: Homework Statement Obtain an expression for: phi(p^3) and phi(P^n) Homework Equations phi(p^2)=p(p-1) The Attempt at a Solution The factors of P^n are 1 and P. Any number less than P^n that shares a factor must have p as a...

But how would you then mvoe onto p^3? Would p^3 then be p(p(p-1) or am I wrong. If I take what you say which is that If a number is NOT relatively prime to p^2 then it must be divisible by p Then can I apply the same rule to p^3?

Generally, I am stumped by the Phi function. I have found out the pattern but I am having difficulty proving it. Homework Statement i) When p is a prime number, obtain an expression in terms of p for:- ϕ(p²)Homework Equations ϕ(p)=p-1 The Attempt at a Solution ϕ(p)=p-1 so ϕ(5)=4 therefore...