TITLE: Find the remainder of $2^{11}$ by $23$
QUESTION [6 upvotes]: My attempt was:
By Fermat's little theorem:
$$2^{22} \equiv 1 \pmod{23}$$
$$(2^{11})^2 \equiv 1 \pmod{23}$$
I checked with my calculator the remainder is actually $1$. However, I wonder if I can take the square root on both sides of congruence. Any idea?  
Thanks,

REPLY [4 votes]: Hint $\rm\ \bmod\ 23\!:\ \ 2 \equiv 5^{\large2}\,  \Rightarrow\, 2^{\large11} \equiv  5^{\large 22} \equiv 1\ $ by Fermat's little Theorem.
See Euler's Criterion and quadratic reciprocity to understand what happens generally, and see the Remark here for the analog with higher power residues.
Regarding square-roots, $\rm\ x^2 = a^2\ \iff\ (x-a)\ (x+a) = 0\ \iff\ x = \pm\: a\ \ $ holds true in any integral domain, i.e. it's true in any ring without zero-divisors. More concretely, in $\rm\ \mathbb Z/p\:,\: $ we have prime $\rm\ p\ |\ (x-a)\ (x+a)\ \Rightarrow\ p\ |\ x-a\ $ or $\rm\ p\ |\ x+a\:,\: $ so $\rm\ x \equiv \pm\: a\ \ (mod\ p)\:.$