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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.

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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

$$x^2-a=x^2-b^2=(x-b)(x+b)=0\pmod p\Longleftrightarrow$$

$$ x=b\pmod p\,\,or\,\,x=-b\pmod p$$

And now you only have to justify why the two solutions above are actually different.

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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?

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    $\begingroup$ when $b\neq -b$ (+1) $\endgroup$ – Belgi Oct 11 '12 at 3:59
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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.

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