<s>[INST] For which $n$ is $2^{2x+2} \equiv 2\pmod n \quad\text{and}\quad 2^{2x+2} \equiv 4\pmod {n-1} $ We begin with 
\begin{align*}
2(2^p+n^2-1-2n) \equiv 0\pmod n \quad&\text{and}\quad 2(2^p+n^2-1-2n) \equiv 0\pmod {n-1} \\
\end{align*}
Which we reduce to 
\begin{align*}
2^p\equiv1\pmod{\frac n{2}} \quad&\text{and}\quad 2^{p-1}\equiv1\pmod{\frac{n-1}{4}}.
\end{align*}
We are looking to prove that $p$ cannot be odd, considering that $n \ge 3$
So we let $p=2x+1$ (looking to prove by contradiction)
\begin{align*}
 2^{2x+2} \equiv 2\pmod n \quad&\text{and}\quad 2^{2x+2} \equiv 4\pmod {n-1} \\
\end{align*}
I know that $a^b \equiv a\pmod b$, so I am assuming that I should go somewhere in that direction. And one more thing, $$\frac{2(2^{p} + n^{2} - 1 - 2n)}{n(n-1)} = s$$ where $s$ is an integer and $s \ge 3$. I have reason to believe that the only time that $p$ can be odd is when $n=2, s=2^p-1$
Also all 3 variables must be positive integers ($s,n,p$)
EDIT: It seems that there are solutions even when $n$ does not equal $2$, but the question still remains, for which $n$ can it hold true
 [/INST] EDIT: For a counterexample, try $p=11$ and $n=23$ We have $2(2^p+n^2−1−2n)=5060=10 \cdot 23 \cdot 22$.
INCORRECT ANSWER:
I think I found a counterexample. Set $p = 3$ and $n = 7$. Then, $2(2^p+n^2−1−2n) = 2(8+49-1-14) = 2(42) = 84$. This is divisible by both $n = 7$ and $n-1 = 6$, but $p = 3$ is odd.
</s>