# Euler's totient function applied to higher power triples

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.

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.

I modified my program and have a few more results.

Unless my program has a bug there are no solutions for $$n = 5, a,b,c \leq 4000$$ or $$n = 4, a,b,c \leq 10000$$

For $$n = 3$$: There are 180 solutions for $$a < b < c \leq 10000$$.

With regards to the theory about sharing prime factors: There are cases where none, 2, or 3 of $$a,b,c$$ share at least one prime factor.

Additionally, there are some cases with $$a < b < c$$ and some with $$a < c < b$$.

This is a non-comprehensive list of my results: $$\phi(16^3)+\phi(46^3)=\phi(45^3)$$ $$\phi(33^3)+\phi(231^3)=\phi(242^3)$$ $$\phi(90^3)+\phi(222^3)=\phi(228^3)$$ $$\phi(107^3)+\phi(354^3)=\phi(251^3)$$ $$\phi(360^3)+\phi(888^3)=\phi(912^3)$$ $$\phi(527^3)+\phi(700^3)=\phi(723^3)$$ $$\phi(1530^3)+\phi(1692^3)=\phi(1956^3)$$

For $$n = 2$$ where $$a,b,c$$ are not a Pythagorean triple: There are a 547 solutions with $$a,b,c \leq 200$$ With $$a < b$$ there was actually at least one example for every value of a from $$a = 1$$ to $$a = 100$$.

• Thanks. At least now we know that examples can be found. I find it interesting that, unlike the case of Pythagorean triples, this triple has no factor common to all three numbers. Which raises the question if we let $n=2$ but skip the requirement that the triple be Pythagorean, might we find many more solutions in the squares? Commented Sep 21, 2018 at 1:20
• I'm working on program improvements out of curiosity. I'll try to update this answer if I can find more or can see any patterns. I was surprised I found an answer with fairly low a,b,c values. Commented Sep 21, 2018 at 1:38
• I can post all of my values which satisfy the equation if someone suggests an appropriate place to post them Commented Sep 21, 2018 at 3:33