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 $\phi(2m)=\phi(m)$.
2026-03-27 12:34:27.1774614867
A question about Euler's totient function
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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 $n = (p-1)m$.