Hi, I have some thoughts about the problem

hello

I am doing a community wiki answer compiling the main results we have gotten so far.

Unfortunately, they seem very basic and far from answering the question

My latest idea is that latex.codecogs.com/gif.latex?\phi(n^2-1)=4\phi\left(\frac{n-1}{4}\right)\‌​phi\left(\frac{n+1}{2}\right) or latex.codecogs.com/gif.latex?\phi(n^2-1)=4\phi\left(\frac{n-1}{2}\right)\‌​phi\left(\frac{n+1}{4}\right) for odd n, so if we can show that n is coprime to both of these values then the map x^(1/n) is unique which forces 3\equiv -5

yeah, they do

You can use MathJaX on chat, see here

Maybe it is an open problem?

have you tried googling?

I also thought about $\varphi(n^2-1)$, but the opportunities I thought about happen so early they must have been completely shattered by the computations.

I don't know what to google, to be honest

yeah, google sucks at math, but that's the downfall of any search engine

You can show, though, what I asserted earlier, which, now that it renders nicer, is $$ \phi(n^2-1)=4\phi\left(\frac{n-1}4\right)\phi\left(\frac{n+1}2\right)\lor 4\phi\left(\frac{n-1}2\right)\phi\left(\frac{n+1}4\right) $$

I see, they must be coprime up to a factor of $2$

yeah, and, as I noted earlier through mod 16, that factor of 2 is $2^3$

For even $n$, you have the much simpler identity $\phi(n^2-1)=\phi(n-1)\phi(n+1)$

$\mod 9$, I showed that the $k$ in $3^n+5^n=k(n^2-1)$ must be equal to $9k'-(-1)^m$, $m$ being $n/3$

So we have $3^{3m}+5^{3m}=(9k'-(-1)^m)(9m^2-1)$

I am trying many modulos, particularly the factors of $3^u+5^v$ for $0\leqslant u,\,v\leqslant 3$

The thing with mods is that they often only narrow down solutions instead of saying there can't be any

Well, it is something, but maybe we can get a contradiction

@TimRatigan have you made more computations?