Rational and irrational fractions over finite fileds

Question

I've been told that over the field $\mathbb{F}_7$ the square root of $2$ is actually $3$.

How come? Why does it happen?

Answer 1

First recall that $x$ is a square-root of an element $y$, if we have $x^2 = y$.

$\mathbb{F}_7$ is the set $\{0\pmod7,1\pmod7,2\pmod7,3\pmod7,4\pmod7,5\pmod7,6\pmod7\}$ with the operations $+$ and $\times$. Note that in this set $$(3 \pmod 7)^2 \equiv 3^2 \pmod7 \equiv 9 \pmod 7 \equiv 2 \pmod7$$ Hence, we get that $3 \pmod7$ is a square-root of $2 \pmod7$. Also, note that $$(4 \pmod7)^2 \equiv 4^2 \pmod 7 \equiv 16 \pmod 7 \equiv 2 \pmod7$$ Hence, the two square-roots of $2 \pmod7$ are $3 \pmod7$ and $4 \pmod7$.