<s>[INST] $\forall a\ge 2, \exists n\ge 1$ such that $2n+1\mid a+n!$ Let $a\ge 2$ be an integer. Prove that $\exists n\ge 1$ such that $$2n+1\mid a+n! \text{ or }a-n!$$
I think we'll construct a solution for $n$ as a function of $a$. But I have no idea what that function looks like. I just took care of some trivial cases.
First if $a$ is a factorial number i.e. $a=k!$ then set $n=k$
If $a$ is of the form $k!+2k+1$ then set $n=k$ and observe that $$a-n!=2k+1$$
Note that if $a\equiv 1$ or $2$ $\pmod 3$ then set $n=1$,so we only have to deal with multiples of $3$ now .
 [/INST] Note that if $2n+1$ is prime, then by Wilson's theorem:
$$-1 \equiv (2n)!\equiv (-1)^{n}(n!)^2 \mod (2n+1)$$
Thus $(n!)^2 \equiv (-1)^{n+1} \mod (2n+1)$
Thus if $2n+1$ is a prime of the form $1\text{mod}$ $ 4$ then $(n)!^2 \equiv -1 \mod(2n+1)$ (also some thanks to John Omielan).
Claim: For all $a \geq 2$, $a^2+1$ contains primes of the form $1 \mod 4$.
Proof:
By quadratic reciprocity, it contains no primes of the form $3\mod 4$. Hence, if it contains no primes of the form $1 \mod 4$ then it is a power of $2$. Thus we must have $a^2+1 = 2^{j}$ with $j \geq 2$ for some $a$. But $a^2+1 \equiv 1 \text{ or }2 \mod 4$, which is impossible.

Let $p_{a}$ be a prime $1 \mod 4$ that divides $a^2+1$. Note that $((\frac{p_{a}-1}{2})!)^2 \equiv -1 \mod p_{a}$ and, therefore;
$$p_{a}| a+\left(\frac{p_{a}-1}{2}\right)! \text{ or }a-\left(\frac{p_{a}-1}{2}\right)!  $$
$$\Leftrightarrow \left(a+\left(\frac{p_{a}-1}{2}\right)!\right)\left(a-\left(\frac{p_{a}-1}{2}\right)!\right) \equiv 0 \mod p_{a}$$
$$\Leftrightarrow a^2+1 \equiv 0 \mod p_{a}.$$
The last statement is true. Thus $n = \frac{p_{a}-1}{2}$ does the trick.
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