Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

3 Answers 3

up vote 1 down vote accepted

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.

share|improve this answer

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?

share|improve this answer
    
when $b\neq -b$ (+1) –  Belgi Oct 11 '12 at 3:59

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.