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Clearly, I'm missing something or misunderstanding something.

I've worked through the same question featured here. I was able to prove the following statement:

Show that if $p$ is an odd prime and $a$ is a positive integer not divisible by p, then the congruence $x^2 \equiv a \pmod{p}$ has either no solution or exactly two incongruent solutions.

But somehow I'm able to come up with counterexamples?

For example, $3$ is an odd prime. $1$ is a positive integer that is not divisible by $3$. Let $a = 1$, $p = 3$. Then $1^2 \equiv 1 \pmod{3}$; $(-1)^2 \equiv 1 \pmod{3}$; $2^2 \equiv 1 \pmod{3}$; and $(-2)^2 \equiv 1 \pmod{3}$; which violates? my conclusion of there being exactly two solutions for $x$ given an $a$ and $p$ as here there appears to be four solutions.

Any idea where my misunderstanding is? Thanks!

Xunnamius
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    $-1\equiv 2 \pmod 3$ and $1\equiv -2\pmod 3$. – lulu Feb 03 '20 at 11:01
  • They say "(in)congruent solutions", i.e. solutions are considered up to congruence, so any congruent solutions are considered to be the same; equivalently, we are working with congruence (equivalence) classes $,a+p\Bbb Z,,$ i.e. elements of the quotient ring $,\Bbb Z/p\ \ \ $ – Bill Dubuque Feb 03 '20 at 19:55

2 Answers2

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Your misunderstanding is that $-1\equiv2$ and $-2\equiv1\pmod3$, so they’re the same solutions.

J. W. Tanner
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  • Ah! So when the problem asks for "exactly two incongruent solutions," we're not taking all integers $x$ and then squaring and applying$\bmod{3}$, but the solutions 1, -1, 2, -2 are compared to each other$\bmod{3}$ (so we're only talking in $Z_3$, as Peter said below)? My explanation isn't well formed I bet, but I think my understanding is getting better. Or to put it another way: -1 is 2, -2 is 1 in in $Z_3$. Thanks! – Xunnamius Feb 03 '20 at 11:20
  • Yes; otherwise we could count $\pm4$, $\pm5$, etc. as "other" solutions – J. W. Tanner Feb 03 '20 at 11:32
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If $\ p\ $ is a prime number, the ring $\mathbb Z_p$ is a field, hence the poylnomial $$x^2-a$$ has at most two distinct root. If $a\ne 0$, a double root is impossible because $$(x-b)^2=x^2-2bx+b^2$$ can only be of the form $x^2-a$, if $b$ is $0$. This completes the proof.

Peter
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