Is it known whether there is a prime $p=4k+1$ such that $k!+1$ is divisible by $p$?
(I conjectured that such primes don't exist, but couldn't prove it.)
Is it known whether there is a prime $p=4k+1$ such that $k!+1$ is divisible by $p$? (I conjectured that such primes don't exist, but couldn't prove it.) 


There are no such primes $p$. Write $p=4k+1$ as $a^2+b^2$ with $a$ odd and $b$ even, and by changing the sign of $a$ if necessary suppose that $a\equiv 1 \pmod 4$. Note that $a$ is uniquely defined. Gauss showed that (see, for example, Binomial coefficients and Jacobi sums for references, and proofs of this and other similar congruences)
$$
\binom{2k}{k} \equiv 2a \pmod p.
$$
Also, by Wilson's theorem we know that $(2k!)^2\equiv (4k)! \equiv 1 \pmod p$. Using this, and squaring Gauss's congruence, we get
$$
4a^2 (k!)^4 \equiv 1 \pmod{p}.
$$
If now $k! \equiv 1 \pmod p$ then we conclude that $4a^2 + 1\equiv 0 \pmod p$. 

