I've been working my way through the mathematics presented in this question: Pythagorean triples that "survive" Euler's totient function concerning Pythagorean triples $a^2+b^2=c^2$ for which it is also the case that $\phi(a^2)+\phi(b^2)=\phi(c^2)$, where $\phi$ is Euler's totient function.

Ignoring the occurrence of exponents, there are many triples $x+y=z$ for which $\phi(x)+\phi(y)\ne \phi(z)$ e.g. $5,7,12$, and conversely many triples $x+y\ne z$ for which $\phi(x)+\phi(y)=\phi(z)$ e.g. $3,7,16$. So there seems to be nothing special about choosing triples whose members are squares other than the limiting condition it places on the triples being looked at, and the interesting answers it affords.

Fermat's Last Theorem establishes that there are no natural number triples $a,b,c$ such that $a^n+b^n=c^n$ for $n>2$. In light of the cited question, I am curious: **Are there any triples such that** $$\phi(a^n)+\phi(b^n)=\phi(c^n)$$ for $n>2$. Of greatest interest would be triples that are non-trivial in the sense that $0<a<b<c$.

I tried applying the logic in the cited question, using numbers sharing common prime factors, but I became a bit overwhelmed. By reference to the specific example of that question, it is certainly the case that $\phi(90^3)+\phi(120^3)\ne \phi(150^3)$. So the simple expedient of choosing triples whose members have common factors does not necessarily yield examples. Furthermore, I suspect that it is unlikely that $a,b,c$ can each be prime numbers, as the relationship $a^{n-1}(a-1)+b^{n-1}(b-1)=c^{n-1}(c-1)$ looks unpromising (but I would enjoy being surprised).

I don't have the programming skills to examine this question computationally, so I am asking for any help or insight the community can provide, either by way of examples or proofs.