Question: let $k$ is postive integer, $p$ is prime number, such $p=3k+1$,$r<p$ be postive integer, such $2^{k+1}\equiv r\pmod p$,and $r\not\equiv 4,5\pmod 6$. show that $$p\nmid k!+1$$.

it is well known $$\left(\dfrac{p-1}{2}\right)!^2+1\equiv 0\pmod p$$ where $p\equiv 1\pmod 4$,for my problem it semm hard to prove it.Thanks

Question: let $k$ be a positive integer, $p$ a prime number, such that $p=3k+1$,  $r<p$ be a positive integer, such that $2^{k+1}\equiv r\pmod p$, and $r\not\equiv 4,5\pmod 6$. Show that $$p\nmid k!+1.$$

It is well known $$\left(\dfrac{p-1}{2}\right)!^2+1\equiv 0\pmod p$$ where $p\equiv 1\pmod 4$, for my problem it seems hard to prove it.

Question: let$k$ is postive integer,and $p$ be prime number, such $p=3k+1$,$r<p$ be postive integers,and such $2^{k+1}\equiv r\pmod p$,and $r\not\equiv 4,5\pmod 6$. show that $$p\nmid k!+1$$

it is well known $$\left(\dfrac{p-1}{2}\right)!^2+1\equiv 0\pmod p$$ where $p\equiv 1\pmod 4$,for my problem it semm hard to prove it.Thanks

Question: let  $k$ is postive integer, $p$ is prime number, such $p=3k+1$,$r<p$ be postive integer, such $2^{k+1}\equiv r\pmod p$,and $r\not\equiv 4,5\pmod 6$. show that $$p\nmid k!+1$$.

it is well known $$\left(\dfrac{p-1}{2}\right)!^2+1\equiv 0\pmod p$$ where $p\equiv 1\pmod 4$,for my problem it semm hard to prove it.Thanks

Question: let $p$ be prime number, and $p=3k+1$,$k,r$ be postive integers,and such $2^{k+1}\equiv r\pmod p$,and $r\not\equiv 4,5\pmod 6$. show that $$p\nmid k!+1=\left(\dfrac{p-1}{3}\right)!+1$$

it is well known $$\left(\dfrac{p-1}{2}\right)!^2+1\equiv 0\pmod p$$ where $p\equiv 1\pmod 4$,for my problem it semm hard to prove it.Thanks

Question: let$k$ is postive integer,and $p$ be prime number, such $p=3k+1$,$r<p$ be postive integers,and such $2^{k+1}\equiv r\pmod p$,and $r\not\equiv 4,5\pmod 6$. show that $$p\nmid k!+1$$

it is well known $$\left(\dfrac{p-1}{2}\right)!^2+1\equiv 0\pmod p$$ where $p\equiv 1\pmod 4$,for my problem it semm hard to prove it.Thanks