TITLE: A question about Euler's totient function
QUESTION [5 upvotes]: Prove that for every natural number $m$, there exists a natural number $n$ such that
$$\phi(n)=\phi(n+m)$$

For odd numbers $m$, we can choose $n=m$ and use the identity $\phi(2m)=\phi(m)$.

REPLY [6 votes]: The provided method for odd numbers can be concisely generalized.  Choose $m$, and let $p$ be the least prime that does not divide $m$, which implies $\phi(pm) = (p-1)\phi(m)$.  Now since the prime factors $p-1$ are all prime factors of $m$ we must have $\phi((p-1)m) = (p-1)\phi(m)$.  Thus, you can choose $n = (p-1)m$.