How far apart can be solutions of $\varphi(m)=\varphi(n)$?

Question

In other words, I was thinking if $\varphi(m)=\varphi(n)$, can $m$ and $n$ be arbitrarily far apart?

That is, is it true that for every $w \in \mathbb N$, there exists $m,n \in \mathbb N$ such that $\varphi(m)=\varphi(n)$ and $|m-n|\geq w$?

Is this much known about Euler´s totient?

Or easily deducible?

Answer 1

Yes, they can be arbitrarily far apart.

Find two numbers $m_0\neq n_0$ with $\varphi(m_0)= \varphi(n_0)$ (like $3$ and $4$). Then for any prime $p$ that divides neither $m_0$ nor $n_0$, we have $$ \varphi(pm_0) = \varphi(pn_0) $$ There are arbitrarily large such primes, meaning $pm_0$ and $pn_0$ can be arbitrarily far apart. For instance, $\varphi(3\cdot 1\,000\,003) = \varphi(4\cdot 1\,000\,003) = 2\,000\,006$, yet the two arguments are over a million apart.

Answer 2

For an arbitrarily large prime number of the form $p=2^k+1$, $\phi(p)=\phi(2^{2k})=2^k$

The two numbers are $2^{2k}-(2^k+1)=2^k-1$ apart, which can be arbitrarily large, assuming only that there are an infinite number of primes of the form $p=2^k+1$