There are infinitely many $n$ such that $\phi(n)=\phi(n-\phi(n))$ (for example $3\cdot 2^k$ for any $k$).

Let $A_>$ be the set of $n$ for which $\phi(n)>\phi(n-\phi(n))$ and $A_<$ the set of $n$ for which $\phi(n)<\phi(n-\phi(n))$. The sequence $A_<$ begins\[30,60,66,120,\ldots.\]Grytczuk, Luca, and Wotowicz [GLW01] proved that $A_>$ has lower density at least $0.54$ and that $A_<$ is infinite.

Indeed, it is easy to check that, for any $k\geq 1$, if $n=15\cdot 2^k$ then $\phi(n)=4\cdot 2^k$ and $\phi(n-\phi(n))=5\cdot 2^{k}$, and so $n\in A_{<}$.

Luca and Pomerance [LuPo02] proved that $A_>$ has density $1$ - in fact, for any function $f(n)=o(n)$, we have $\phi(n)>\phi(n-\phi(n))+f(n)$ for almost all $n$. They also prove that, for any constant $c>0$, the inequality $\phi(n)< c\phi(n-\phi(n))$ holds for infinitely many $n$.


This is mentioned in problem B42 of Guy's collection [Gu04].

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This page was last edited 06 October 2025.