In the past I tried to get different variations of the so-called Chinese hypothesis, see this Wikipedia (a disproven conjecture).
Today I wanted to combine in an artificious way also Wilson-Lagrange theorem to prove next claim.
Claim. If $n$ is prime then $$((n-1)!+3)^n\equiv 2\text{ mod }n.\tag{1}$$
Sketch of the proof. It is obvious from Wilson-Lagrange $LHS\equiv 2\cdot 2^{\varphi(n)}\text{ mod }n$, where $\varphi(n)$ is the Euler's totient.$\square$
Question. I don't know if it is obvious that:
If an integer $n$ satisfies $((n-1)!+3)^n\equiv 2\text{ mod }n$ then $n$ is a prime number.
Can you find such easy proof for previous statement, or well refute it with a conunterexample? Thanks in advance.
Now I'm stuck about how to prove it, if is right. I tested the first few integers with this code
mod(((n-1)!+3)^n, n), from n=1 to 100
One counterexample is $n=2929 = 29\times 101$.
Note that $2928! = 0 \mod 2929$, so the given statement is equivalent to $3^{2929} = 2 \mod{2929}$. This can be checked by computer, although checking by hand is feasible by use of the Chinese Remainder Theorem and Fermat's Little Theorem.