Great common divisor and prime number.

Question

Let $p$ be a prime number do we have :

$$\gcd\left(p!-p^{2},1+2p\right)=^?1$$

Except the Wilson's theorem I haven't a clue to show this while I don't find a counter-example for the 25 first primes .

How to (dis)prove it ?

Answer 1

Suppose there is a prime $q$ that divides $\gcd\left(p!-p^{2},1+2p\right)$. Then $q\mid p!-p^{2}$ and $q\mid 1+2p$

If

• $q\leq p$ then $q\mid p!$ and thus $q\mid p$ so $q\mid 1$ a contradiction.
• $q>p$ then from $1+2p = kq$ we get $k=1$ or $k=2$. Clearly $k=2$ is impossible so $k=1$ and $q=2p+1$. By Willson theorem we have $$p!^2 \equiv_q (-1)^{p+1} \equiv_q 1\implies p! \equiv_q \pm 1$$ Now we have two subcases

  • $p! \equiv_q 1$ then $$q\mid p^2-1=(p-1)(p+1)$$ and thus $2p+1 \mid p-1$ or $2p+1\mid p+1$ whic is impossible in both cases since $2p+1$ is bigger natural than $p-1$ and $p+1$.

  • $p! \equiv_q -1$ then $$q\mid p^2+1$$ which is impossible by Fermat little theorem since $q\equiv_4 3$.