A corollary Zsigmondy's Theorem leads to the following congruence (one can look to $(24)$),$\phi(a^n+b^n)=0\mod n$ whenever $a, b$ are coprime and $n \neq 2$ and $(a,b)\neq(1,1)$. (Here $\phi$ is the Euler totient function.) Is there any characterization of when $\phi(a^n+b^n+c^n)=0\bmod n$ assuming that $a,b,c$ are coprime integers?

1$\begingroup$ I have no idea how your initial claim follows from Zsigmondy's Theorem (could you explain?). It is nevertheless true for $(a,b)\neq(1,1)$ though, as we can easily see $ab^{1}$ has order $2n$ modulo $a^n+b^n$ (as the least $k$ for which $(ab^{1})^k\equiv\pm 1$ is $n$ for size reasons). This argument sadly doesn't generalize to three numbers. $\endgroup$ – Wojowu Apr 23 at 22:30

$\begingroup$ you may check this and see (24) $\endgroup$ – user147204 Apr 23 at 22:38

$\begingroup$ @Wojowu, Have you checked (24) in mathworld for looking how the titled claims follows ? I have linked it in the question $\endgroup$ – user147204 Apr 26 at 2:19