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.)
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1 Answer
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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 $4a^2 + 1\equiv 0 \pmod p$.
Since $4a^2+1$ is $1\pmod 4$ and at most $4p$, we must have $4a^2+1 =p = (2a)^2 +1^2$. But by the (essential) uniqueness of writing $p$ as a sum of two squares this forces $a=1$ (and so $p=5$). For $p=5$ we verify directly the claim.
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$\begingroup$ Where can I find a proof of that congruence? $\endgroup$ May 7, 2014 at 21:42
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$\begingroup$ @user50519: I added a reference to a paper that discusses this. $\endgroup$– LuciaMay 7, 2014 at 21:48
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$\begingroup$ I do not understand the end of your argument: (essential) uniqueness forces $a=1$. Example: $a=-3$ leads to $p=37$, $a=5$ to $101$. What additional information forces $a=1$? $\endgroup$ May 11, 2015 at 15:15
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$\begingroup$ @RolandBacher: Note throughout that $a^2+b^2=p$. Note that for example $37=1^2+6^2$, so you can't take $a=-3$ there. (What's being used at the end is that there's essentially only one way to write a prime $p\equiv 1\pmod 4$ as a sum of two squares.) $\endgroup$– LuciaMay 11, 2015 at 15:19
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$\begingroup$ I see. You need to remember the second sentence ($a$ odd) which I did not. Thanks. $\endgroup$ May 11, 2015 at 15:27