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I need a proof for the following:

Suppose that $p$ is an odd prime. If $(a, p) = 1$, then $x^2 = a \pmod p$ either has exactly $2$ solutions or has no solutions within $\textrm{crs}/p$.

I can come up with a lot of examples that work, but I am having trouble with the proof.

user26486
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Dillon
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    The polynomial $x^2-a$ over the finite field $\mathbb F_p$ is separable. In other words look up the condition for a polynomial to have a multiple root. You'll see for $p$ odd the quadratic never does. – Gregory Grant Apr 13 '15 at 15:33
  • Just to be clear: by $(a, p) = 1$ you are referring to the Legendre symbol and not $\gcd(a, p) = 1$, right? –  Apr 13 '15 at 17:47
  • @AlonsodelArte No -- that would imply there is always at least one solution. – user26486 Apr 13 '15 at 18:23
  • @user31415 So that's a yes to GCD and a no to Legendre? –  Apr 13 '15 at 18:59
  • @AlonsodelArte right. The $\gcd(a,p)=1$ is there because if we had $p\mid a$, then $a\equiv 0\pmod {p}$ and so there is a unique (not two, not zero) solution $x\equiv 0$. – user26486 Apr 13 '15 at 19:00
  • @user31415 Then I encourage Dillon to rewrite as "$\gcd(a, p) = 1$." I do know that $(a/p)$ would be more conclusively the Legendre symbol, but it doesn't rule out the possibility he meant to type a slash but typed a comma instead. –  Apr 13 '15 at 19:01
  • @AlonsodelArte $(a,b)$ to denote $\gcd(a,b)$ is conventional and usual. Whereas I've never seen anyone denote $(a,b)=\left(\frac{a}{b}\right)$. – user26486 Apr 13 '15 at 19:03
  • @user31415 Not anyone in a peer-reviewed journal. But I wouldn't make that assumption about a random person on a website. –  Apr 13 '15 at 19:16

1 Answers1

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If it has a root $\,b^2\equiv a\,$ then $\,0\equiv x^2\!-\!b^2\equiv (x\!-\!b)(x\!+\!b).\,$ Then prime $\,p\mid (x\!-\!b)(x\!+\!b)\,$ so $\,p\mid x\!-\!b\,$ or $\,x\!+\!b,\,$ so $\,x\equiv b\,$ or $\,-b,\,$ and $\,-b\not\equiv b,\,$ else $\, 2b\equiv 0,\,$ contra $\,(p,2)=1=(p,b^2)$

Remark $\ $ More generally a commutative ring $\ne 0$ is an integral domain $(a,b\ne 0\,\Rightarrow, ab\ne 0)$ iff nonzero polynomials over it have no more roots than their degree, as is easily shown via an inductive proof using the Factor Theorem, e.g. see the BiFactor Theorem.

Community
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Bill Dubuque
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  • @use It shows: if there is a solution $,x\equiv b,$ then there are exactly two solutions $,x\equiv \pm b\pmod p.\ \ $ – Bill Dubuque Apr 13 '15 at 16:08