Diophantine equation $p^{k}=lk+1$ with $p$ prime and $l,k$ integers

Question

Solve the equation : $$p^k=kl+1, $$ with $p$ a prime number and $k,l\ge 1$ two integers.

I know that $(p,k,l)=(3,1,2)$ is a solution, but can we find all solutions to the equation ?

Answer 1

COMMENT.- Putting $k=1$ one has $p=l+1$ so there are infinitely many solutions.

With $k=2$ we have the equation $p^2=2l+1$ then since $p=4m\pm1$ there are again infinitely many solutions with $l=8m^2\pm4m$.

It is possible pursue this way for greater values of $k$.

Answer 2

An easy set of solutions will be found for all $p\ge 5$ because those primes have the form $6m\pm 1$. For $k=6$, and any prime $\ge 5$, $p^k=p^6=(6m\pm 1)^6=(36m^2\pm 12m+1)^3=(6M+1)^3=6l+1=kl+1$.