Find all $n\in \mathbb{Z^+}, \ \phi (3n)=\phi (4n)=\phi (6n)$

Question

We denote with $\phi$ Euler's Phi Function.

We want to find all $n\in \mathbb{Z^+}: \ \phi (3n)=\phi (4n)=\phi (6n).$

Answer: Let $n=2^k3^lm$ : $k,l\in \mathbb{N}, \ m\in \mathbb{Z^+},\ \text{gcd}(m,2)=\text{gcd}(m,3)=1$. Then we have:

1. $\phi (3n)=\phi (4n) \implies \phi(2^k)3^l=2^k\phi(3^l) $
   • If $k,l>0$ we have contradiction.
   • If $k=0,l>0\implies 3^l=\phi(3^l)\implies l=0.$
   • If $l=0,k>0\implies 2^k=\phi(2^k) \implies k=0$.

So, $\phi (3n)=\phi (4n) \iff n\in A=\{m\in \mathbb{Z^+}: \text{gcd}(m,6)=1\}$.

2. $\phi (4n)=\phi (6n)\implies \phi(3^l)=3^l\implies l=0 $.

So, $\phi (4n)=\phi (6n) \iff n\in B=\{2^km\in \mathbb{Z^+}: \text{gcd}(m,6)=1, k\in \mathbb{N}\} \supseteq A$.

Finally, $\ \phi (3n)=\phi (4n)=\phi (6n) \iff n \in A \cap B=A \iff n=m \in \mathbb{Z^+}:\text{gcd}(m,6)=1.$

Is this proof completely right? And, moreover, is there another way to proove it?

Thank you.

Answer 1

We have $\phi(3)=\phi(4)=\phi(6)=2$, and for any $n$ coprime to $2$, $3$, $6$, hence to $6$, we have, because $\phi$ is multiplicative $$ \phi(3n)=\phi(4n)=\phi(6n)=2\phi(n). $$ If $n$ is divisible by $2$, then $\phi(4n)>\phi(3n)$, so the above equality does not hold. Similarly for the other cases.

Answer 2

As you begin, let $n = 2^a 3^b m$ with $\gcd (m, 6) = 1$. The given conditions can then be rewritten as $\phi (2^a 3^{b+1}) = \phi (2^{a+2} 3^b) = \phi (2^{a+1} 3^{b+1})$, i.e. $\phi (2^a) \phi(3^{b+1}) = \phi (2^{a+2}) \phi (3^b) = \phi (2^{a+1}) \phi(3^{b+1})$.

As boring as it may be, you need now examine several cases.

1. If $a=b=0$ then the above is $2 = 2 = 2$, which means that all numbers with $\gcd (m, 6) = 1$ are good.

2. If $a=0$ and $b>0$ then the given equality implies $2 \cdot 3^b = 2 \cdot 2 \cdot 3^{b-1} = 2 \cdot 2 \cdot 3^b $ which is easily seen to be impossible.

3. If $a>0$ and $b=0$ then the given equality implies $2^{a-1} \cdot 2 = 2^{a+1} = 2^a \cdot 2$, again impossible.

4. If $a>0$ and $b>0$ then the given equality implies $2^{a-1} \cdot 2 \cdot 3^b = 2^{a+1} \cdot 2 \cdot 3^{b-1} = 2^a \cdot 2 \cdot 3^b$, again impossible.

Therefore, the solution is $\{m \in \Bbb Z \mid \gcd (m, 6) = 1\}$.

Answer 3

The shortest approach I can see uses the following simple principle: $\frac{\phi(n)}n$ is determined by just the set of primes dividing $n$ (and not their exponents). This allows us to make deductions without any explicit calculations regarding exponents.

Suppose that $n$ satisfies the required condition. Then since $\phi(3n) = \phi(6n)$ is non-zero, $\frac{\phi(3n)}{3n} \ne \frac{\phi(6n)}{6n}$, so $3n$ and $6n$ have distinct sets of prime divisors $\implies n$ is odd.

Since $n$ is odd, $\phi(4n) = 2 \phi(n)$, which by assumption is equal to $\phi(3n)$. Thus $\frac{\phi(n)}{n} \ne \frac{\phi(3n)}{3n}$, so $n$ and $3n$ have distinct sets of prime divisors $\implies n$ is not divisible by $3$.

Thus $n$ must be relatively prime to $6$.

Finally, this necessary condition is sufficient since $\phi(3) = \phi(4) = \phi(6) = 2$ and $(n,6)=1$ implies $\phi(3n) = \phi(4n) = \phi(6n) = 2\phi(n).$