Show that prime 1223 satisfies
$$1223 \mid (2^{611}− 1)$$
Hint: Use Euler’s Criterion.
We remind ourselves of Euler's Criterion (7.5).
Let $p$ be an odd prime such that $p \not \mid a$. Then $a$ is a quadratic residue of $p$ if and only if $a^{\frac{p-1}{2}} \equiv 1 \pmod p$.
We will use Proposition (7.15):
Let $p$ be an odd prime then
$$ (\frac{2}{p}) = \begin{cases} 1 & \text{ if } p \equiv \pm 1 \pmod 8 \\ -1 & \text{ if } p \equiv \pm 3 \pmod 8 \end{cases} $$
Since the prime $1223 \equiv -1 \pmod 8$, we have $(\frac{2}{p})=1$, by Proposition (7.15).
This means, by Euler's Criterion (7.5)
$$ 2^{\frac{1223-1}{2}} \equiv 2^{611} \equiv 1 \pmod {1223} $$
Re-arranging gives us the desired result
$$ 1223 \mid (2^{611}-1) $$