On an iterated arithmetic function problem of Erdos and Graham
Stefan Steinerberger
TL;DR
The paper resolves the r=2 case of Erdős and Graham's iterated arithmetic function problem by proving the equivalence to the Diophantine equation $φ(m)+φ(m+φ(m))=m$ and fully solving the resulting structure. It yields six infinite families given by $n=2^{}old or ll ll$ in ${1,3,5,7,35,47}$, and further shows any non-power-of-two solution must arise from primes with a rare totient property, reducing to the congruence $φ(q)=(2/3)(q+1)$. This leads to the parametrization $q=6m+5$ and $p=(4q+1)/3=8m+7$, with primes $p$ satisfying $φ(6m+5)=4m+4$; only $m=0$ and $m=5$ yield such primes ($p=7,47$), giving explicit $g$-orbits, while a computer search finds no further primes up to $10^{10}$. The result suggests a rigid, low-dimensional structure in the dynamics of the iterated totient map and indicates that the six small seeds may exhaust all small solutions, with any additional ones tied to extremely rare prime seeds.
Abstract
Erdős and Graham define $g(n) = n + φ(n)$ and the iterated application $g_k(n) = g(g_{k-1}(n))$. They ask for solutions of $g_{k+r}(n) = 2 g_{k}(n)$ and observe $g_{k+2}(10) = 2 g_{k}(10)$ and $g_{k+2}(94) = 2 g_{k}(94)$. We show that understanding the case $r = 2$ is equivalent to understanding all solutions of the equation $φ(n) + φ(n + φ(n)) = n$ and find the explicit solutions $ n = 2^{\ell} \cdot \left\{1,3,5,7,35,47\right\}$. This list of solutions is possibly complete: any other solution derives from a number $n=2^{\ell} p$ where $p \geq 10^{10}$ is a prime satisfying $φ((3p-1)/4) = (p+1)/2$. Primes with this property seem to be very rare and maybe no such prime exists.