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Question: Define a function: \(\phi :N \to N\) as follows : \(\phi (1) = 1,\phi ({P^n}) = {P^{n - 1}}(P - 1)\)...

Define a function: ϕ:N→N\phi :N \to N as follows : ϕ(1)=1,ϕ(Pn)=Pn−1(P−1)\phi (1) = 1,\phi ({P^n}) = {P^{n - 1}}(P - 1) , if P is prime and n∈Nn \in N and ϕ(mn)=ϕ(m)ϕ(n)\phi (mn) = \phi (m)\phi (n) if m and n are relatively prime numbers.
ϕ(8n+4)\phi (8n + 4) where n∈Nn \in Nis equal to
(E). 2ϕ(4n+2)2\phi (4n + 2)
(F). ϕ(2n+1)\phi (2n + 1)
(G). 2ϕ(2n+1)2\phi (2n + 1)
(H). 4ϕ(2n+1)4\phi (2n + 1)

Explanation

Solution

A function is a special relation between two sets. Here the function is defined as ϕ:N→N\phi :N \to Nsuch that ϕ(1)=1,ϕ(Pn)=Pn−1(P−1)\phi (1) = 1,\phi ({P^n}) = {P^{n - 1}}(P - 1) holds for prime numbers. So we need to evaluate ϕ(8n+4)\phi (8n + 4)using the given relation and provide ϕ(mn)=ϕ(m)ϕ(n)\phi (mn) = \phi (m)\phi (n) holds for all natural numbers that are relatively prime.

Complete step by step solution:
Now,
ϕ(8n+4)\phi (8n + 4) can be written by taking 4 common as:
= \phi (4(2n) + 4(1)) \\\ = \phi (4(2n + 1)) \\\
So, now we can take 4 as a scalar and comparing it with ϕ(mn)=ϕ(m)ϕ(n)\phi (mn) = \phi (m)\phi (n) we will rewrite the equation in the simplified form as:
=ϕ(4)ϕ(2n+1)= \phi (4)\phi (2n + 1) .
=ϕ(22)ϕ(2n+1)= \phi ({2^2})\phi (2n + 1)
Now, clearly we know that 2 and 2n+12n + 1 are relatively prime numbers. So, now we will use
ϕ(Pn)=Pn−1(P−1)\phi ({P^n}) = {P^{n - 1}}(P - 1). With P=2P = 2 , we will get the expansion as:
\phi ({P^n}) = {P^{n - 1}}(P - 1) \\\ \Rightarrow \phi ({2^n}) = {2^{n - 1}}(2 - 1) \\\
Now putting this we get:
ϕ(22)ϕ(2n+1) =22−1(2−1)ϕ(2n+1) =21(1)ϕ(2n+1) =2ϕ(2n+1) \phi ({2^2})\phi (2n + 1) \\\ = {2^{2 - 1}}(2 - 1)\phi (2n + 1) \\\ = {2^1}(1)\phi (2n + 1) \\\ = 2\phi (2n + 1) \\\
Therefore, the correct answer is option C.

Note: A prime number is a number whose only factors are 1 and the number itself. Two numbers are relatively prime when the highest common factor between those two numbers is 1. 2 is the only even prime number. So 2 and any odd number will always be relatively prime since the only number common between them will be 1.