$ x^n = (y - 1)\frac {y^{n + 1} - 1}{y - 1}$ Other : $ \gcd(y - 1,\frac {y^{n + 1} - 1}{y - 1}) = \gcd(y - 1,n + 1)$ We prove that : $ \gcd(y - 1,n + 1) = 1$ Suppose $ p|\gcd(y - 1,n + 1)$ Imply that $ p|x^n\rightarrow p|x$ so $ p|\gcd(x,n + 1) = 1$ So p must be equal 1. So $ \gcd(y - 1,n + 1) = 1$ Imply that exist $ (a,b)\in N$ $ y - 1 = a^n,\frac {y^{n + 1} - 1}{y - 1} = b^n,x = ab$ and $ \gcd(a,b) = 1$ $ y^n + y^{n - 1} + ... + 1 = b^n$ If $ n\geq 2$ we has $ y^n < \frac {y^{n + 1} - 1}{y - 1} < (y + 1)^n$ so it has no solution. So $ n - 1$ $ x = y^2 - 1$ Because $ \gcd(x,2) = 1$ so the solution is $ y = 2k,x = 4k^2 - 1$

Peter wrote: Find all $ (x,y,n) \in {\mathbb{N}}^3$ such that $ \gcd(x, n + 1) = 1$ and $ x^{n} + 1 = y^{n + 1}$. $ y^{n+1}-x^{n}=1$ Taking clue from Mihailescu's theorem, it is clear that the only solutions are when n=1, meaning x is odd. $ y^{2}=x+1$ So y is even. Let y=2k $ 4k^{2}-1=x$ $ (2k+1)(2k-1)=x$ But this has infinitely many solutions, since any value k gives a unique value of x, with (x,n+1)=1, and with $ (x,y,n)$ being integers above 0. So, all ordered triplets of the form $ (4k^{2}-1,2k,1)$ work.

The equation $ x^n + 1 = y^{n + 1}$ can be written as $ \frac{{x^n }}{{y - 1}} = (y^n - 1) + (y^{n - 1} - 1) + \cdots + (y - 1) + (n + 1)$ Now if $ (\frac{{x^n }}{{y - 1}},y - 1) = d$ then $ d|(n + 1)$ and therefore $ d = 1$ since $ d|x^n$. But $ \frac{{x^n }}{{y - 1}} \cdot y - 1 = x^n$so $ \frac{{x^n }}{{y - 1}} = a^n ,y - 1 = b^n$ and then $ y^n + ny^{n - 1} + \cdots + 1 = (y + 1)^n \le a^n = y^n + \cdots + y +1$ Which implies $ n = 1$, hence $ a = y + 1$, $ x = ab = (y - 1)(y + 1) = y^2 - 1$. Since $ n + 1 = 2$ is even $ x$ must be odd and then $ y = 2m$ is even. So $ (4m^2 - 1,2m,1)$ are the desired triples.

by LTE lemma $ y-1 = A^n $ mod4: $ n|v_2(y-1) $