The following equation does not seem to have any non-trivial integer solutions for prime $p\ge5$ and constant $r$: $$p x (1 + x) r = y^p - 1$$
Could you please help with proving of assumption?
The solutions $x=0$,$r=0$ and $x=-1$ with $y=1$ are treated as trivial.
This is Thue equation so this means it has finite set of solutions.
What I have found so far which may help in proving:
- (x,y)=1
- (($y-1$),($1+...+y^{p-1}$))=$1$ or $p$
There are some other findings which I am not sure will help during the solving.
And also counterexamples are also something I was trying to find.
EDITED:
According to Michael Burr comment and taking into account the fact that the $x(x+1)$ is even:
$$y\equiv1\pmod{2p}$$
Pick any prime $p$, and any number $k\geq 1$. Define $y=2pk+1$.
Then by the binomial expansion one sees that $$r=\dfrac{y^p-1}{2p}\in \mathbb{N}$$ Then $x=1$ is a solution to your equation.
Instead of any $k$, pick $k=3$. Again $y=2pk+1=6p+1$. Then $$r=\dfrac{y^p-1}{6p}\in \mathbb{N}$$
This time $x=2$ is a solution.
It seems that you can build solutions by finding a suitable $k$ and $ r $ such that $$1+4\dfrac{(2pk+1)^p-1}{pr}$$ is a perfect square.
My method is first find a perfect square congruent to $1$ mod $4$, then adjust $k$ and $r$.