Let $p$ be an odd prime and $a$ be an integer with $\gcd(a, p) = 1$. Show that $x^2 - a \equiv 0 \mod p$ has either $0$ or $2$ solutions modulo $p$
I am clueless with this one. Hints please.
For $\,p=2\,$ the claim fails, as $\,x^2-1=x^2+1=(x+1)^2=0\pmod 2\,$ has one unique solution.
If $\,p\,$ is odd and $\,x^2-a=0\pmod p\,$ has a solution $\,b\,$, then
$$ x=b\pmod p\,\,or\,\,x=-b\pmod p$$
And now you only have to justify why the two solutions above are actually different.
If $x^2-a=0 \mod p$ has some solution $b$, it means that $b^2=a$, and hence you original question becomes
$$x^2-b^2 \equiv 0 \mod p$$
Can you prove now that this equation has exactly two solutions?
Hint $\ $ In a field, $\rm\ x_2^2 = x_1^2 \iff 0 = x_2^2-x_1^2 = (x_2\!-x_1\!)\,(x_2\!+x_1\!)\iff x_2 = \pm\, x_1$
Remark $\ $ Generally a nonzero polynomial over a domain has no more roots than its degree.