Exercise (7.2).14

Assuming 2 is a primitive root of the prime 101, find the square roots of $14 \pmod {101}$.

As per the previous exercise, even powers of the primitive root 2 are congruent to quadratic residues modulo 101.

The following table shows the first even powers of 2.

n	2^n mod 101
2	4
4	16
6	64
8	54
10	14

We can read off that

$$ 2^{10} \equiv (2^5)^2 \equiv (\pm 32)^2 \equiv 14 \pmod {101} $$

And so the square roots of 14 modulo 101 are $32 \pmod {101}$ and $69 \pmod {101}$.