Find all pairs of integers $(n,k)$ that satisfy $(n+1)^k - 1 = n!$

It is easy to see that (1,1) and (2,1) are solutions by inspection, but how do we prove that these are the only solutions? After sometime, I also saw that $(4,2)$ was also a solution, as it wasn't so obvious.

I was thinking that maybe $n! + 1$ cannot be a perfect power of a number? Though I also can't prove it. I also tried binomial theorem but still cant see anything.

  • 1
    $\begingroup$ $4!+1=5^2$, $5!+1=11^2$. $\endgroup$
    – lulu
    Nov 11, 2018 at 13:49
  • $\begingroup$ @lulu can we find relation between $a,b$ where $a!+1=b^2$? $\endgroup$
    – Fawad
    Nov 11, 2018 at 13:57
  • $\begingroup$ @Fawad Maybe...we also have $7!+1=71^2$, but there are no other examples for $a≤200$. $\endgroup$
    – lulu
    Nov 11, 2018 at 13:59
  • $\begingroup$ @Fawad Turns out that it is an open problem, conjectured that the three examples I wrote out are all of them. Brocard's Problem. $\endgroup$
    – lulu
    Nov 11, 2018 at 14:03


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