# When is $\phi(a^n+b^n+c^n)=0\mod n$?

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?

• 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. – Wojowu Apr 23 at 22:30
• you may check this and see (24) – user147204 Apr 23 at 22:38
• @Wojowu, Have you checked (24) in mathworld for looking how the titled claims follows ? I have linked it in the question – user147204 Apr 26 at 2:19