Question

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

Answer 1

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

Complete step by step solution:
Now,
$\phi (8n + 4)$ 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 $\phi (mn) = \phi (m)\phi (n)$ we will rewrite the equation in the simplified form as:
$ = \phi (4)\phi (2n + 1)$ .
$ = \phi ({2^2})\phi (2n + 1)$
Now, clearly we know that 2 and $2n + 1$ are relatively prime numbers. So, now we will use
$\phi ({P^n}) = {P^{n - 1}}(P - 1)$. With $P = 2$ , 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:
$\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.