is it true that $P(x)=x-3$ the only solution to this problem?

Question

Question: Find all monic polynomial $P(x)$ with integer coefficients such that there exists a natural number $L$, satisfying: $$p\ | \ 2\times (P(p)!)+1$$ for every prime number $p$ greater than $L$

The first polynomial that I found was $P(x)=x-3$, since: $$2\times (p-3)! \equiv (2-p)\times(p-3)! \equiv -(p-2)! \equiv (p-1)! \equiv-1 \ (mod \ p)$$ (choosing $L=3$)

However I cannot find any other polynomial that satisfies the question. Is it true that $P(x)=x-3$ the only solution to this problem?

(Sorry, English is my second language)

Answer 1

If $\deg P\geq 2$, eventually $P(p)\geq p$, which makes

$$2(P(p))!+1\equiv 2\cdot 0+1\equiv 1\bmod p.$$

So, $P$ is of degree $1$, and as it is monic, it is $x-a$ for some positive integer $a$ ($a>0$ as otherwise we run into the aforementioned problem).

Using that

$$-1\equiv (p-1)!=(p-a)!(p-a+1)\cdots(p-1)\equiv (p-a)!(-1)^{a-1}(a-1)!,$$

can you finish from here?