Question

Prove that there do not exist natural numbers $x$ and $y,$ with $x>1,$ such that $\frac{x^7-1}{x-1}=y^5+1$.

Answer 1

Simple factorization gives $y^5=x(x^3+1)(x^2+x+1)$.
The three factors on the right are mutually coprime and hence they all have to be fifth powers.
In particular, $x=r^5$ for some integer $r$.
This implies $x^3+1=r^{15}+1$, which is not a fifth power unless $r=-1$ or $r=0.$
$\therefore$ there are no solutions to the given equation.