Factorial-exponent equality

A roughly stated solution is presented here.

if $(n,k)$ is a valid pair, then

$(n+1)^k \equiv 1 \ (mod t) \ , \ \forall 2 \leq t \leq n \ (*)$

immediately one notices that $n+1=p$ should be a prime, otherwise there would be a $2 \leq t \leq n$ s.t. $gcd(t,n+1)\neq 1$ and consequently $*$ cannot hold true for that particular $t$ .

then we realise $k|lcm(\{\phi(t)\}_{t=2}^{t=n})=1$ or $k|lcm(\{\phi(t)\}_{t=3}^{t=n})=2$ , when we assume $p^k \equiv 1 \ (mod 2)$ . Therefore, possibilities are $k=1,2$ .

with $k=1$ , we have $(n+1)-1=n! \implies n=n! \implies n=1,2$ .

When $k=2$ , we have

$p^2 \equiv 1 \ (mod t) \ \forall 2 \leq t \leq p-1 \implies t|(p-1)(p+1) \ \forall 2 \leq t \leq$

according to Bertrand's postulation, there should be a prime $q$ such that

$\frac{p+1}{2} < q < p$

when $p \leq 5$ .Therefore $q \nmid (p-1)(p+1)$ and, for this $q$ , equation $*$ would not be satisfied. the only primes $p$ , for which there would be no such $q$ , are $2,3,5$ . One can plug the primes, together with $k=2$ in $*$ and easily check if they constitute a valid pair.

The only possible pairs are (1,1), (2,1), (4,2)