Solve the following equations for $x, (x ∈ N^+) (a)\ 2φ(x) = x; (b)\ 3φ(x) = x; (c)\ 4φ(x) = x.$

Question

Solve the following equations for $x, (x ∈ N^+)$

$$ (a)\ 2φ(x) = x;$$ $$(b)\ 3φ(x) = x;$$ $$(c)\ 4φ(x) = x.$$

Can anyone help me with this, or give me some hint?

Answer 1

$(a) 2φ(x)=x$

$x = 2^2$

$φ(x)=2^1*(2-1)=2$

$2φ(x)=2*2=2^2=x \Rightarrow 2φ(x)=x$

$(b) 3φ(x)=x$

$x=2*3^3=54$

$φ(x)=φ(2*3^3)=φ(2)φ(3^3)=18$

$3φ(x)=3*18=54=x \Rightarrow 3φ(x)=x$

$(c) 4φ(x)=x$

$x=2^3*3*4^4$

$φ(x)= φ(2^3*3*4^4)=2^3*3*4^4$

$4φ(x)=4*2^3*3*4^4=2^3*3*4^4=x\Rightarrow 4φ(x)=x$

Answer 2

HINT:

For $$x=\prod_\limits{i} p_i^{a_1} =p_1^{a_1}\cdot p_2^{a_2}\cdot p_3^{a_3} \ldots p_n^{a_n}$$ where $p_i$'s are the prime factors of $x$ raised to respective powers $a_i$ in the prime factorisation of $x$.

use the fact that $$\phi(x)=x\left(1-\frac{1}{p_1}\right)\left(1-\frac{1}{p_2}\right)\left(1-\frac{1}{p_3}\right)\ldots \left(1-\frac{1}{p_n}\right)$$