Why does $x^2 \equiv 2 \pmod{5}$ have no solution? Can someone explain why $x^2 \equiv 2 \pmod{5}$ has no solution other than trial and error? 
So I've shown that modulo 5, we have $0^2 \equiv 0, 1^2 \equiv 1, 2^2 \equiv 4, 3^2 \equiv 4, 4^2 \equiv 1$ and I'm seeing that it doesn't look like there is a solution but what is the actual reason? Since I can't go through infinitely many numbers. 

You have already shown what you need. Every integer is of the form $0, 1, 2, 3$, or $4$ modulo $5$, and  $0^2$ mod $5 = 0$, $1^2$ mod $5 = 1$, $2^2$ mod $5 = 4$, $3^2$ mod $5 = 4$, and $4^2$ mod $5 = 1$, so we know that there is no integer $x$ for which $x^2 \equiv 2(mod\;5)$ i.e. $x^2(mod\;5) \neq 2$ for all integers $x$.
