 # Understand the problem

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

##### Source of the problem
Singapore MO 2008
Number Theory
Medium
##### Suggested Book
An Excursion in Mathematics

Do you really need a hint? Try it first!

Note that $n+1$ has to be a prime, because any proper prime divisor of $n+1$ would divide $n!$ too.
Say $n+1=p$. Note that, for large enough $p$, both $\frac{p-1}{2}$ and $2$ are factors of $(p-1)!$. Thus, the factor $(p-1)$ occurs twice in the expansion of $(p-1)!$.
Prove that, if $(p-1)^2|p^k-1$ then $p-1|k$.
Hint 3 implies that $p-1\le k$. Hence $p^{p-1}-1\le p^k-1=(p-1)!$. This is obviously false. Hence, $p$ has to be small enough to avoid this situation. It is avoided precisely when $p\in\{2,3,5\}$. This corresponds to $n\in\{1,2,4\}$. The equations to be solved are $2^k=2, 3^k=3$ and $5^k=25$. Hence, the solutions are $\{(1,1),(2,1),(4,2)\}$.

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