Trivial by Mihailescu's Theorem

t0rajir0u wrote: Mihailescu's Theorem What is it?

Mihailescu is the one who proved Catalan's conjecture.

ZetaX wrote: Mihailescu is the one who proved Catalan's conjecture. But what is Mihailescu's Theorem ?

Well, proving a conjecture leads often to them be named after you... Oh, and there is also google

Ok! It here http://mathworld.wolfram.com/CatalansConjecture.html But above problem has level is Olympiad?

i smel here short listed problem 2000. find all natural numbers such:$a^{m}+1=(a+1)^{n}$. and also there is generalisation for this $a^{m}+1|(a^{k}+1)^{n}$. i will post my solution using elementry number theory.

Both are trivial after using Zsygmoni (and it is elementary)

ZetaX wrote: Zsygmoni What is it? again

Sorry, it's spelled "Zsigmondy", see http://mathworld.wolfram.com/ZsigmondyTheorem.html .

N.T.TUAN wrote: Find all triples $(x,y,z)$ of positive integers such that $(x+1)^{y+1}+1=(x+2)^{z+1}$. By Mihailescu's Theorem: $\Rightarrow (x,y,z)=(1,2,1)$ By Zsigmondy's Theorem: $\exists$ prime $p/ p|(x+1)^{y+1}+1, p\nmid x+2$ $\Rightarrow 0\equiv (x+2)^{z+1} \pmod{p} (\Rightarrow \Leftarrow)_\blacksquare$ But there are exceptions: $\Rightarrow (x,y,z)=(1,2,1)$

ZetaX wrote: Sorry, it's spelled "Zsigmondy", see http://mathworld.wolfram.com/ZsigmondyTheorem.html . Thank you I need this too