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Ian Mateus
Ian Mateus
14:02
Hi, I have some thoughts about the problem
Tim Ratigan
Tim Ratigan
hello
Ian Mateus
Ian Mateus
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
Tim Ratigan
Tim Ratigan
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
Ian Mateus
Ian Mateus
You can use MathJaX on chat, see here
5
Maybe it is an open problem?
Tim Ratigan
Tim Ratigan
have you tried googling?
Ian Mateus
Ian Mateus
14:09
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
Tim Ratigan
Tim Ratigan
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 $$ \phi(n^2-1)=4\phi\left(\frac{n-1}4\right)\phi\left(\frac{n+1}2\right)\lor 4\phi\left(\frac{n-1}2\right)\phi\left(\frac{n+1}4\right) $$
Ian Mateus
Ian Mateus
I see, they must be coprime up to a factor of $2$
Tim Ratigan
Tim Ratigan
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 $\phi(n^2-1)=\phi(n-1)\phi(n+1)$
Ian Mateus
Ian Mateus
$\mod 9$, 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 $0\leqslant u,\,v\leqslant 3$
Tim Ratigan
Tim Ratigan
The thing with mods is that they often only narrow down solutions instead of saying there can't be any
Ian Mateus
Ian Mateus
14:19
Well, it is something, but maybe we can get a contradiction
Ian Mateus
Ian Mateus
14:49
@TimRatigan have you made more computations?

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