$a^2 = 1\pmod{p}$ Could someone prove if $p$ is a prime number, $a$ can't be nontrivial?
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This is pretty standard: $$a^2-1 \equiv 0 \mod p$$ means $p| (a-1)(a+1)$. Since $p$ is prime, we get either $p|a-1$ or $p|a+1$. Thus $a\equiv 1 \mod p$ or $a \equiv -1 \mod p$. P.S. If you know what a primitive root is, you can try to prove the following generalization: If there is a primitive root mod $n$, then the equation $x^2=1 \mod n$ has only two solutions.... |
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The solution given by N.S. is good. Here is a slightly more "high level" look on it: It is well-known that $\mathbb{Z}_p$ (integers modulo $p$ with addition and multiplication) is a field (all the axioms can be checked directly; to show that $a\ne 0$ is invertible use the fact that $gcd(a,p)=1$ and so $ax+py = 1$ for some integers $x,y$). Now, it is known that if $p(x)$ is a polynomial of degree $n$ over any field, then $p(x)$ has at most $n$ roots; this is so because of $p(x)$ has root $a$ then $(x-a)$ divides $p(x)$ and we can proceed by induction. Now $x^2-1$ is a polynomial of degree 2 over the field $\mathbb{Z}_p$ and so it has at most two roots. We know of the two trivial roots $1,-1$ (-1 being $p-1$) and so there are no other roots. |
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